Mathematical methods in physics

Part I B

Lectures given by Frank Krauss,michaelmas terms 2006 and 2007

at University of Durham

Disclaimer:

This is the second half of part I of the course “Mathematical methods inphysics”. The first half has been given by Prof. V. Khoze. This documentdoes not intend to replace the lecture, it just contains handouts summarizingthe lectures. In the preparation of this lecture, a number of excellent bookshas been used, among them:

• G. G. Stephenson: “Mathematical methods for science students”

• M. L. Boas: “Mathematical methods in phsiycal sciences”

• M. R. Spiegel: “Fourier analysis”

Contents

1 Matrices, Determinants . . . 11.1 Matrices and their properties . . . . . . . . . . . . . . . . . . 1

1.1.1 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.2 Determinants . . . . . . . . . . . . . . . . . . . . . . . 61.1.3 Solving systems of linear equations . . . . . . . . . . . 11

1.2 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . 141.2.1 Similarity Transformations . . . . . . . . . . . . . . . . 141.2.2 Eigenvalues and eigenvectors . . . . . . . . . . . . . . . 19

2 Fourier Series 272.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.1.1 Simple harmonic motion . . . . . . . . . . . . . . . . . 272.1.2 Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.1.3 More frequencies . . . . . . . . . . . . . . . . . . . . . 29

2.2 Trigonometric and Exponential functions . . . . . . . . . . . . 302.2.1 General properties of functions . . . . . . . . . . . . . 302.2.2 Some trigonometry . . . . . . . . . . . . . . . . . . . . 31

2.3 Definition & convergence . . . . . . . . . . . . . . . . . . . . . 362.3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . 362.3.2 The form of the Fourier coefficients . . . . . . . . . . . 402.3.3 Convergence: Dirichlet conditions . . . . . . . . . . . . 412.3.4 Uniform convergence of series . . . . . . . . . . . . . . 422.3.5 The Weierstrass M-test . . . . . . . . . . . . . . . . . . 43

2.4 Simple properties . . . . . . . . . . . . . . . . . . . . . . . . . 442.4.1 Parseval’s identity . . . . . . . . . . . . . . . . . . . . 442.4.2 Sine and Cosine series . . . . . . . . . . . . . . . . . . 442.4.3 Integration and differentiation . . . . . . . . . . . . . . 482.4.4 Complex notation . . . . . . . . . . . . . . . . . . . . . 51

i

2.4.5 Double and multiple Fourier series . . . . . . . . . . . 532.5 Some applications . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.5.1 Full wave rectifier . . . . . . . . . . . . . . . . . . . . . 542.5.2 Quantum mechanical particle in a box . . . . . . . . . 552.5.3 Heat expansion . . . . . . . . . . . . . . . . . . . . . . 582.5.4 Electrostatic potential . . . . . . . . . . . . . . . . . . 60

3 Fourier transforms 633.1 Definitions and properties . . . . . . . . . . . . . . . . . . . . 63

3.1.1 The Fourier integral . . . . . . . . . . . . . . . . . . . 633.1.2 Detour: The δ-function . . . . . . . . . . . . . . . . . . 673.1.3 Fourier transforms . . . . . . . . . . . . . . . . . . . . 70

3.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 763.2.1 Finite wave train . . . . . . . . . . . . . . . . . . . . . 763.2.2 Infinite string . . . . . . . . . . . . . . . . . . . . . . . 77

3.3 Green functions . . . . . . . . . . . . . . . . . . . . . . . . . . 803.3.1 Green function for the driven harmonic oscillator . . . 80

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Chapter 1

Matrices, Determinants,Eigenvalues and Eigenvectorsand all that

1.1 Matrices and their properties

1.1.1 Matrices

Systems of Linear Equations

At many occasions in physics and mathematics, systems of linear equationsemerge. These are, in general, m equations for n unknowns, i.e.

a11x1 + a12x2 + . . . + a1nxn = b1a21x1 + a22x2 + . . . + a2nxn = b2

......

. . ....

...am1x1 + am2x2 + . . . + amnxn = bm .

(1.1)

Introducing the matrix Awithm rows, one for each equation, and n columns,one for each unknown,

A =

⎛⎜⎜⎜⎝

a11 a12 . . . a1n

a21 a22 . . . a2n...

.... . .

...am1 am2 . . . amn ,

⎞⎟⎟⎟⎠ (1.2)

1

and the n-component vectors �x and �b, allows to represent equation systemsof this type like

A�x = �b . (1.3)

Alternatively, the same set of equations can also be written as

n∑k=1

alkxk =

n∑k=1

xkalk = bl , (1.4)

where l ∈ [1, m]. Using Einsteins convention of summing over re-peated indices this can be written as

alkxk = xkalk = bl . (1.5)

In the case considered here, a matrix with n columns and m rows, the matrixis said to be of type (m, n) or just a (m, n)-matrix. The entries aij arecalled the elements of the matrix. In the notation here, the first indexof the elements (i) denotes the row, whereas the second one (j) denotesthe column. Clearly, if there are more unknowns than equations, n > mthe system is under-determined. If, in contrast, m > n, the system isover-determined and it is possible that no solution can be constructed. Ingeneral, in this case the presence of viable solutions indicates that pairs ofrows are multiples of each other or that one row can be re-expressed througha sum of other rows. In other words, in such a case, the rows are not linearlyindependent. If m = n, the matrix is called quadratic of order n, and inthe general case of linearly independent rows and columns, a solution canuniquely be found. If all entries aij with i �= j are zero, and entries withi = j are non-zero, the matrix is called diagonal, and a solution is simple.If such a diagonal matrix has all diagonal entries equal to one, i.e. aij = δij ,it is called the unit matrix, often denoted by 1.A standard method to solve such a system is called Gauss elimination. Inthis method, basically the first row/equation is employed to eliminate thefirst unknown, the (potentially altered) second row/equation is used for thesecond unknown and so on. Thereby, the allowed operations here are:

• Interchanging two rows;

• Multiplying/dividing a row with/by a constant;

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• Adding and/or subtracting multiples of two rows.

Example:

• To see how this works, consider the three equations

2x − z = 26x + 5y + 3z = 72x − y = 4 .

(1.6)

To alleviate things, in a first step the first two rows are interchanged,leaving

2x − z = 22x − y = 46x + 5y + 3z = 7 .

(1.7)

Eliminating the first unknown with the first equation implies subtract-ing the first equation from the second, and subtracting three times thefirst equation from the third, yielding

−y + z = 25y + 6z = 1 .

(1.8)

The second unknown is eliminated by adding five times the first row ofthe emerging system to the second one, such that

11z = 11 and z = 1 . (1.9)

Reinserting then gives y = −1 and x = 3/2.

Properties of matrices

Matrices, such as A in the previous section, allow a number of operations,which will be briefly listed here:

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• Transposition:A matrix can be transposed by interchanging rows and columns, i.e.

AT = {aij}T = {aji} . (1.10)

If AT = A the matrix is called symmetric, if AT = −A the matrix isantisymmetric.

• Adjoint Matrix:Given a matrix A with complex elements, its adjoint matrix A† is givenby transposing it and complex conjugation of each element:

A† = {aij}† ={a∗ji

}. (1.11)

Matrices which satisfy

A† = A (1.12)

are called Hermitian or self-adjoint. Such matrices play a centralrole in Quantum Mechanics, because their eigenvalues (see below) areguaranteed to be real. Also, their diagonal elements are real numbers.

• Addition:Two matrices are added to yield another matrix,

A + B = C , (1.13)

where the elements of C are given by

cij = aij + bij . (1.14)

This obviously assumes that all three matrices are of the same type,i.e. have the same number of rows and columns.

• Multiplication with a number:A matrix can be multiplied by a number by multiplying each elementwith this number:

cA = Ac = C = {cij} = {caij} . (1.15)

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• Multiplication:Two matrices are multiplied to yield another matrix,

AB = C , (1.16)

where the elements of C are given by

cij = aikbkj = bkjaik , (1.17)

i.e. to obtain the elements in the ith row and the jth column of C,the elements in the ith row of A are multiplied one-by-one with theelements in the jth column of B, and the products are summed over.This automatically implies that if A and B are of type (n, m) and(k, l), respectively, then m = k and C is a (n, l)-matrix.

It is quite clear from the expression above that in general

AB = aikbkj = bkjaik �= bikakj = akjbik = BA . (1.18)

• Rank:The rank of a matrix is given by a non-negative integer number, whichis given in the following way: Consider a (n, m)-matrix and take thecolumns as vectors of dimension m. The rank of the matrix is thengiven by the maximal number of linearly independent vectors in thisset. The rank of a matrix does not change under:

– permutation of the columns,

– multiplication of a column with any number c �= 0,

– addition of an arbitrary multiple of one row to any other row,

– transposition of the matrix,

– and, hence, permutation of the rows.

In addition, there are some properties which are defined for quadratic matri-ces only:

• Trace:The trace of a matrix is defined as the sum of its diagonal elements:

Tr(A) = aii . (1.19)

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• Inverse:The inverse of a matrix A, A−1, is defined such that

AA−1 = A−1A = 1 (1.20)

or, in components,

aija−1jk = δik . (1.21)

• Determinant:See next section.

More complicated properties

Having defined the inverse and transposed (Hermitian conjugate) of a matrixit is worth to state a few more important properties:

(AB)−1 = B−1A−1 (1.22)

and

(AB)T = BT AT

(AB)† = B†A† . (1.23)

In order to see this, let us start with the transposed:

(AB)T = (aijbjk)ki = ajibkj = bTjkaTji = BT AT . (1.24)

For the inverse consider the product

1 = (AB)−1(AB) = B−1A−1AB = B−1B = 1 . (1.25)

1.1.2 Determinants

Definition

One of the most important properties of a quadratic matrix is its determi-nant, a well-defined, unique real or complex number. It is denoted by

det(A) = |A| =

∣∣∣∣∣∣∣∣∣

a11 a12 . . . a1n

a21 a22 . . . a2n...

.... . .

...an1 an2 . . . ann

∣∣∣∣∣∣∣∣∣=∑Π

(−1)j(Π) a1i1a2i2 . . . anin .

(1.26)

6

Here, the sum stretches over all n! possible permutations Π of the numbers1 to n, and j(Π) is the sign of the respective permutation. In practise thisimplies that each product consists of n elements in such a way that it containsexactly one elements per row and column.To underline this, just think of any matrix as being related to a “checker-board” of the same size filled with + or − signs,⎛

⎜⎜⎜⎝+ − + . . .− + − . . .+ − + . . ....

......

. . .

⎞⎟⎟⎟⎠ . (1.27)

For calculating the determinant of a n×n-matrix, all possible combinationsmust be summed of products with n matrix element, such that each prod-uct contains exactly one matrix element per column and row. The overallsign of such a product is then determined by the sign of the correspondingcheckerboard entries, which are all identical.

Example:

•

det

(a bc d

)=

∣∣∣∣ a bc d

∣∣∣∣ = ad− bc . (1.28)

Properties

Before discussing how a determinant can be calculated using the method ofLaplace, the properties of determinants will briefly be listed:

• Permutation of rows or columns does not change the absolute valueof the determinant but only its sign;

• multiplying a row by a number results in multiplying the determinantwith the same number;

• adding a multiple of one row to another row does not change the de-terminant;

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• the determinant is zero if the rows are not linearly independent;

• the determinant is invariant under transposition.

There is yet another important property namely the behavior of the deter-minant under multiplication:

det(AB) = det(A)det(B) . (1.29)

This implies, in particular,

det(AA−1) = det(1) = det(A)det(A−1) (1.30)

and thus

det(A−1) =1

det(A). (1.31)

Calculating determinants

In principle, determinants of 2×2- or 3×3-matrices can be calculated straight-forwardly: ∣∣∣∣ a11 a12

a21 a22

∣∣∣∣ = a11a22 − a12a21 (1.32)

and∣∣∣∣∣∣a11 a12 a13

a21 a22 a23

a31 a32 a33

∣∣∣∣∣∣ = a11

∣∣∣∣ a22 a23

a32 a33

∣∣∣∣ + a12

∣∣∣∣ a23 a21

a33 a31

∣∣∣∣ + a13

∣∣∣∣ a21 a22

a31 a32

∣∣∣∣= a11[a22a33 − a32a23] + a12[a23a31 − a21a33] + a13[a21a32 − a31a22] .

(1.33)

In the calculation of the latter determinant, Laplace’s method already hasbeen indicated. The idea here is to use either a row or column (in the caseshown here, it was the first row) of an n× n-matrix, and to multiply each ofthese elements with the determinant of the (n−1)× (n−1)-submatrix whichforms when ignoring the row and column of this element in the original ma-trix. The determinant of this submatrix is sometimes denoted as the adjointof the element aij , denoted by Aij . This method is called development

8

of the determinant according to the mth row or column, formallyspeaking

detA =∑j

aijAij =∑i

aijAij , (1.34)

where the index i (j) is fixed to the row (column) in question.

Example:

• ∣∣∣∣∣∣1 −5 27 3 42 1 5

∣∣∣∣∣∣ = 2

∣∣∣∣ 7 32 1

∣∣∣∣− 4

∣∣∣∣ 1 −52 1

∣∣∣∣ + 5

∣∣∣∣ 1 −57 3

∣∣∣∣= 2 · 1 − 4 · 11 + 5 · 38 = 148 . (1.35)

• ∣∣∣∣∣∣∣∣2 9 9 42 −3 12 84 8 3 −51 2 6 4

∣∣∣∣∣∣∣∣c2→c2−c3+c4=

∣∣∣∣∣∣∣∣2 4 9 42 −7 12 84 0 3 −51 0 6 4

∣∣∣∣∣∣∣∣c3→c3/3

= 3

∣∣∣∣∣∣∣∣2 4 3 42 −7 4 84 0 1 −51 0 2 4

∣∣∣∣∣∣∣∣c1↔c2= −3

∣∣∣∣∣∣∣∣4 2 3 4−7 2 4 80 4 1 −50 1 2 4

∣∣∣∣∣∣∣∣= −3

⎧⎨⎩4

∣∣∣∣∣∣2 4 84 1 −51 2 4

∣∣∣∣∣∣− (−7)

∣∣∣∣∣∣2 3 44 1 −51 2 4

∣∣∣∣∣∣− 0 + 0

⎫⎬⎭

= −3

⎧⎨⎩4 · 0 + 7

∣∣∣∣∣∣2 3 44 1 −51 2 4

∣∣∣∣∣∣⎫⎬⎭ r1→r1−r3= −21

∣∣∣∣∣∣1 1 04 1 −51 2 4

∣∣∣∣∣∣= −21

∣∣∣∣ 1 −52 4

∣∣∣∣ + 21

∣∣∣∣ 4 −51 4

∣∣∣∣= −21(4 + 10) − 21(−5 − 16) = 147 . (1.36)

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Calculating the inverse of a matrix

The inverse of a matrix is defined through

AA−1 = 1 or aija−1jk = δik . (1.37)

For the determination of the inverse of a n × n-matrix, there are actuallythree methods, which are related to each other:

1. Direct calculation:The school book method is to calculate it directly through

A−1 =1

detA

⎛⎜⎜⎜⎝

A11 A12 . . . A1n

A21 A22 . . . A2n...

.... . .

...An1 An2 . . . Ann

⎞⎟⎟⎟⎠T

, (1.38)

where the Aij are the adjoints, i.e. determinants of (n − 1) × (n − 1)-matrices.

Examples:

• 2 × 2-matrix:(3 14 2

)−1

=1

2

(2 −4−1 3

)T

=1

2

(2 −1−4 3

). (1.39)

Obviously(3 14 2

)· 1

2

(2 −1−4 3

)=

1

2

(2 00 2

)= 1 . (1.40)

• 3 × 3-matrix:⎛⎝ 3 2 1

1 0 24 1 3

⎞⎠

−1

=1

5

⎛⎝ −2 −5 4

5 5 −51 5 −2

⎞⎠ . (1.41)

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2. Solving a system of linear equations:Going back to the definition,

AX = AA−1 = 1 (1.42)

it is clear that for the n2 entries of the unknown matrix X (which issupposed to become the inverse A−1) there are n2 systems of linearequations. Solving them yields the desired result.

3. “Back of the envelope” for small matrices:This method is also known as Gauss-Jordan method. The idea hereis to have the original matrix and the unit matrix of the same size nextto each other, say, the original matrix to the left and the unit matrixto the right. Then, on both matrices an identical series of steps isperformed such that the original matrix is manipulated such that theunit matrix results. This is exemplified in Tab. 1.1.

At this point it should be stressed that in general the existence of the inverseof a matrix is not guaranteed. This can be seen directly from the occurrenceof the determinant in the denominator in the first calculation method. Italso shows under which circumstances an inverse does not exist, namely ifthe rank of the matrix is smaller than its actual dimension.

1.1.3 Solving systems of linear equations

This allows for the solution of systems of linear equations, i.e. of

A�x = �b or aijxj = bi . (1.43)

If all elements of�b equal zero, bi = 0 then this is called a homogeneous systemof linear equations.

Behavior of the solutions

A number of points are worth noting:

1. If the system A�x = �b is homogeneous, bi = 0, then there always isa trivial solution, namely the null vector, i.e. xi = 0. The relevantquestion therefore is whether there are other, non-trivial solutions.

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Example:⎛⎝ 3 2 1

1 0 24 1 3

⎞⎠

⎛⎝ 1 0 0

0 1 00 0 1

⎞⎠ exchange 1st and 2nd row

⎛⎝ 1 0 2

3 2 14 1 3

⎞⎠

⎛⎝ 0 1 0

1 0 00 0 1

⎞⎠ subtract multiples of 1st row

⎛⎝ 1 0 2

0 2 −50 1 −5

⎞⎠

⎛⎝ 0 1 0

1 −3 00 −4 1

⎞⎠ subtract 3rd row from 2nd row

⎛⎝ 1 0 2

0 1 00 1 −5

⎞⎠

⎛⎝ 0 1 0

1 1 −10 −4 1

⎞⎠ subtract 2nd row from 3rd row

⎛⎝ 1 0 2

0 1 00 0 −5

⎞⎠

⎛⎝ 0 1 0

1 1 −1−1 −5 2

⎞⎠ divide 3rd row by −5

⎛⎝ 1 0 2

0 1 00 0 1

⎞⎠

⎛⎝ 0 1 0

1 1 −11/5 1 −2/5

⎞⎠ subtract twice 3rd row

⎛⎝ 1 0 0

0 1 00 0 1

⎞⎠

⎛⎝ −2/5 −1 4/5

1 1 −11/5 1 −2/5

⎞⎠

Table 1.1: ”Back of the envelope”-inversion of a matrix.

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2. If �x1 and �x2 are linearly independent solutions of the homogeneousproblem, then any linear combination of them is a solution of the ho-mogeneous problem. This follows directly from the linearity.

3. If �xi is a solution of the inhomogeneous problem and �xh is a linearlyindependent solution of the homogeneous problem, then any linear com-bination of �xi and �xh is a solution of the inhomogeneous problem. Thisfollows directly from the linearity.

Let us consider the case where A is quadratic. Then, obviously, multiplyingwith A−1 from the right yields

A−1A�x = �x = A−1�b (1.44)

or

a−1ij ajkxk = δikxk = xi = a−1

ij bj . (1.45)

In case the inverse of A exists, the solution is unambiguous. This is obviouslynot the case, if detA = 0 - in this case not all rows and columns of the matrixare linearly independent, and therefore the system is under-determined.

Cramer’s rule

Another way to solve a system of n equations for n unknowns is based ondeterminants and known as Cramer’s rule. Although, in principle, row reduc-tion is superior in terms of calculation steps for matrices explicitly consistingof numbers, Cramer’s method may be advantageous if the matrix elementsare not numerical but, e.g., functions. To see how this method works, how-ever, numerical examples are best-suited. Therefore, consider the simplestcase, two equations for two unknowns

a11x1 + a12x2 = b1 (1.46)

a21x1 + a22x2 = b2 . (1.47)

Multiplying the first row by a22 and the second one by a12 and subtractingboth yields

x1 =b1a22 − b2a12

a11a22 − a21a12, (1.48)

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and, similarly,

x2 =b2a11 − b1a21

a11a22 − a21a12. (1.49)

Obviously, this is a viable solution only, if the denominator in both cases isdifferent from zero. This can be cast into

x1 =

∣∣∣∣ b1 a12

b2 a22

∣∣∣∣∣∣∣∣ a11 a12

a21 a22

∣∣∣∣, x2 =

∣∣∣∣ a11 b1a21 b2

∣∣∣∣∣∣∣∣ a11 a12

a21 a22

∣∣∣∣. (1.50)

The recipe which determinants to take for this and the general case reads asfollows: Take the matrix of the coefficients, A. Its determinant in all casesyields the denominator. For the solution of xm, replace the mth column ofA, the aim with the inhomogeneous coefficients bi and form the determinantof this new matrix. It yields the numerator.

1.2 Eigenvalues and Eigenvectors

1.2.1 Similarity Transformations

Linear Transformations of vectors

Consider the effect of a matrix A on a vector �x: When correctly multiplied,a new vector �x′ emerges, i.e.

aijxj = x′i (1.51)

or

�x′ = φ(�x) = A�x . (1.52)

Clearly, the transformation law φ here has the following properties:

φ(�x+ �y) = A�x+ A�y = φ(�x) + φ(�y) (1.53)

and

φ(λ�x) = A(λ�x) = λA�x , (1.54)

14

where λ is a real number. These are exactly the properties of a linear trans-formation, see below.In general, linear transformations of a n-dimensional vector, where n maybe infinite, can be represented by n × n matrices. The result of such atransformation is then obtained by multiplying the vector and the matrix (inQuantum Mechanics the lingo will be that an operator acts on a vector).To be more specific and, in fact, more intuitive, consider vectors of dimension3 only, which can be used to specify, e.g., the position or momentum of aparticle or similar. The norm of such a three-vector is given by

||�x|| =√�x2 =

√�x · �x =

√xixi =

√x2i =

√x2

1 + x22 + x2

3 . (1.55)

Clearly, the norm of such a vector is then identical to its length.

Orthogonal and unitary transformations

Linear transformations, acting on a real vector space, that leave the norm of avector unchanged are called Orthogonal Transformations. Such matriceswith real elements which represent transformations of real vectors that leavethe norm invariant, i.e.

�x′ · �x′ = x′iTx′i = (Aijxj)

T (Aikxk) = xjAjiAikxk = xixi (1.56)

have the following important property:

xjAjiAikxk = xixi =⇒ AjiAik = δjk =⇒ AT = A−1 . (1.57)

This, in fact, holds true for all orthogonal transformations.Another important case occurs, if the transformations act on complex vectorspaces, such as Cn. In this case, the norm of a vector is defined through

|�v|2 =n∑i=1

|ai|2 = aia∗i , (1.58)

where the asterisk denotes complex conjugation. In this case, the transforma-tions that leave the norm invariant are called Unitary Transformations,and the corresponding matrices have the property

M † = M−1 . (1.59)

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In both case, however, linear transformations which leave the norm of vectorsinvariant have a determinant equal to ±1.

Example:As an example, consider for a moment rotations of this vector, clearly leavingits length unchanged and merely altering the direction. Such an operationcan be interpreted in two ways: Either the vector actually rotated and theaxes remained fixed, or the vector remained fixed, whereas the axes rotated(in the other direction). So, it should not come as a surprise that the normof vectors remains unchanged.As an example consider counterclockwise rotations by an angle φ around thez-axis. Such transformations are mediated by matrices of the form

Tφ =

⎛⎝ cosφ − sinφ 0

sin φ cos φ 00 0 1

⎞⎠ . (1.60)

It is simple to check that indeed T Tφ = T−1φ and that detTφ = 1.

As a side-product, consider two subsequent rotations around the angles φand ψ around the z-axis. It is well-known that for sequences of rotationsaround the same axis, the angles just add up, such that

Tφ+ψ =

⎛⎝ cos(φ+ ψ) − sin(φ+ ψ) 0

sin(φ+ ψ) cos(φ+ ψ) 00 0 1

⎞⎠

= TφTψ =

⎛⎝ cosφ − sinφ 0

sin φ cosφ 00 0 1

⎞⎠⎛⎝ cosψ − sinψ 0

sinψ cosψ 00 0 1

⎞⎠

=

⎛⎝ cosφ cosψ − sin φ sinψ − cos φ sinψ − sinφ cosψ 0

sin φ cosψ + cosφ sinψ − sin φ sinψ + cos φ cosψ 00 0 1

⎞⎠(1.61)

From this, some trigonometric relations are readily identified, namely

cos(φ+ ψ) = cosφ cosψ − sinφ sinψ

sin(φ+ ψ) = sinφ cosψ + cosφ sinψ . (1.62)

16

Properties of orthogonal and unitary matrices

One of the more interesting properties of orthogonal matrices is that multiply-ing two of them again yields an orthogonal matrix. Consider two orthogonalmatrices O1 and O2 and their product (O1O2). Clearly,

1 =(O1O2

)(O1O2

)−1

= O1O2O−12 O−1

1 . (1.63)

Inserting that for orthogonal matrices OT = O−1 then yields

1 = O1O2OT2 O

T1 =

(O1O2

)(O1O2

)T, (1.64)

where Eq. (1.23) has been used. Therefore,

(O1O2

)−1

=(O1O2

)T, (1.65)

as advertised.Similarly, it can be shown that the product of two unitary matrices again isunitary. To this end, assume to unitary matrices U1 and U2. Obviously,

1 =(U1U2

)(U1U2

)−1

= U1U2U−12 U−1

1 = U1U2U†2 U

†1 . (1.66)

Here the fact that for unitary matrices U † = U−1 has been used. Sinceadjoint is - up to complex conjugation - nearly identical to transposition,

U †2 U

†1 =

(U1U2

)†, (1.67)

cf. Eqs. (1.23,1.24) and therefore

(U1U2

)−1

=(U1U2

)†, (1.68)

fulfilling the definition of unitary matrices.

17

Group theory

Using the findings above, it is quite simple to show that the sets of all orthog-onal or unitary transformations, respectively, which act on the same vectorspace, form a non-abelian group under multiplication. The above reasoningshows closure, i.e. the fact the product of two such orthogonal (unitary)transformations again yields an orthogonal (unitary) transformation. Sincethey are represented by matrices, their associativity,

A(BC) = (AB)C (1.69)

is simple to show. The identiy element in both cases is the unit matrixof corresponding dimension, and the inverse are uniquely defined and againorthogonal or unitary matrices.

Linear Transformations, general definition

Unique maps φ relating two vector spaces V and V′ are called linear mapsif they enjoy the following properties:

φ(�x+ �y) = φ(�x) + φ(�y) ∀ �x, �y ∈ V ; (1.70)

and

φ(λ�x) = λφ(�x) ∀ �x ∈ V and ∀ real numbers λ . (1.71)

If, in addition, for a linear map φ the two vector spaces V and V′ are iden-tical, V = V′ then the map is called Linear Transformation or LinearOperator.

Examples:

• The map φ(x1, x2, x3) = (x1, x2), connecting the R3 with the R2 islinear. But the vector spaces are not identical, therefore, this is not alinear transformation.

• The map φ(x1, x2, x3) = (x1, 1), connecting the R3 with the R2 is notlinear, because

φ(�x+ �y) = (x1 + y1, 1) �= (x1, 1) + (y + 1, 1) = (x1 + y1, 2) . (1.72)

18

From the properties above, it is clear that linear maps φ can be representedby matrices:If dim(V) = m and dim(V′) = n the matrix has dimensions n×m. In orderto see why a matrix is a good representation, it suffices to convince yourselfthat each vector �x ∈ V can be written as a linear combination of the basevectors of V:

�v =

m∑i=1

ai�e(V)i (1.73)

and thus the essence of the linear transformation is in the information of howthe base vectors of V transform under φ. This representation is not quiteunique: Each map V → V′ is represented by a set of matrices, one per com-bination of base vectors. The specific matrix then emerges in the followingway: The kth column of the matrix T is given by the (n-dimensional) vectorin V′ emerging as a result of φ acting on the kth base vector of V:

tik = [φ(�e(V)k ]i = [T�e

(V)k ]i . (1.74)

Thus, switching from one set of base vectors to another base set changes thematrix T representing φ.However, a pair of matrices representing the same transformation but fordifferent choices of base vectors is called equivalent. Formally speaking, achange of base can be realized by another linear transformation, representedby a matrix S. Thus,

TB = S−1B→B

TBSB→B , (1.75)

where the old and new set of bases B and B have been made explicit. Sucha transformation is called a Similarity Transformation.It can be shown that such a similarity transformation leaves both thedeterminant and the trace of a matrix invariant.

1.2.2 Eigenvalues and eigenvectors

Definition

Eigenvalues �λ(i) and eigenvectors λ(i) of a quadratic matrix A are defined by

A�λ(i) = λ(i)�λ(i) . (1.76)

19

In other words: Eigenvectors are those vectors, which, when being subjectedto a matrix A yield a multiple of themselves, where the multiple is the cor-responding eigenvalue.It is important to stress here that, if �λ(i) is an eigenvector, then also anymultiple c�λ(i) with c ∈ R is an eigenvector.The equation above can also be written as

(A− λ(i)1)�λ(i) = 0

(akl − λ(i)δkl)λ(i)l = 0 . (1.77)

This is a linear equation for each i. Solving it through determinants byCramer’s rule would yield only the trivial solution, λ

(i)m = 0, unless the de-

terminant of the coefficients was equal to zero,

det(A− λ(i)1) = 0 . (1.78)

In this case, the individual solutions would expressions of the type 0/0. How-ever, in this case, the rows of the new matrix A − λ(i)1 are not linearly in-dependent, resulting in infinitely many solutions for each value λ(i) - whichis the desired result, since, if any vector is an eigenvector than also realmultiples of this vector is an eigenvector.However, Eq. (1.78) often is called the characteristic equation for A.

Examples:

• Consider

A�λ(i) =

(5 −2−2 2

)(xy

)= λ(i)

(xy

)= λ(i)�λ(i) . (1.79)

Solutions for the eigenvalues are obtained by solving

det(A− λ(i)1) =

∣∣∣∣ 5 − λ −2−2 2 − λ

∣∣∣∣ = λ2 − 7λ+ 6 = 0 . (1.80)

Hence,

λ(1,2) =7 ±√

49 − 24

2=

7 ± 5

2(1.81)

or

λ(1) = 6 and λ(2) = 1 . (1.82)

20

This fixes the eigenvalues. In order to fix the eigenvectors, the followingtwo systems of linear equations must be solved

5x(1) − 2y(1) = 6x(1) 5x(2) − 2y(2) = x(2)

−2x(1) + 2y(1) = 6y(1) −2x(2) + 2y(2) = y(2) (1.83)

for (x(1,2), y(1,2)). Sets of solutions are given by one equation in eachcase, namely

−x(1) = 2y(1) and 2x(2) = y(2) . (1.84)

This shows that the eigenvectors are multiples of

�λ(1) =1√5

( −21

)and �λ(2) =

1√5

(12

), (1.85)

where a the vectors have already been normalized to unit length.

In order to diagonalise the matrix, build a transformation matrix Tfrom the eigenvectors and its inverse,

T =(�λ(1) �λ(2)

)=

1√5

( −2 11 2

)and T−1 =

1√5

( −2 11 2

).(1.86)

Then

T−1MT = · · · =

(6 00 1

)=

(λ(1) 00 λ(2)

), (1.87)

as anticipated.

• As a 3-dimensional example consider the rotation matrix of the exampleabove,

Tφ =

⎛⎝ cosφ − sinφ 0

sin φ cos φ 00 0 1

⎞⎠ . (1.88)

There the characteristic equation reads

0 = (cos φ− λ)2(1 − λ) + sin2 φ(1 − λ)

= (1 − λ)(1 − 2 cosφλ+ λ2) (1.89)

21

leading to

λ(1) = 1 (1.90)

with

�λ(1) = �ez (1.91)

and

λ(2,3) = cosφ±√

cos2 φ− 1 = cos φ± i sinφ . (1.92)

The occurrence of complex numbers is slightly disturbing here, but, ofcourse, this may happen. It should be noted that in this case usuallyalso the eigenvectors become complex.

Diagonalizing a matrix

In the most general case, a matrix M may be diagonalized by invoking alinear transformation represented by a matrix T such that

T−1MT = D , (1.93)

where D has diagonal form. The question is how the transformation matrixand the diagonal matrix can be determined. To this end, write the equationsdetermining the n eigenvalues and eigenvectors in matrix form,

M

⎛⎜⎜⎜⎝

λ(1)1 λ

(2)1 . . . λ

(n)1

λ(1)2 λ

(2)2 . . . λ

(n)2

......

. . ....

λ(1)n λ

(2)n . . . λ

(n)n

⎞⎟⎟⎟⎠

=

⎛⎜⎜⎜⎝

λ(1)1 λ

(2)1 . . . λ

(n)1

λ(1)2 λ

(2)2 . . . λ

(n)2

......

. . ....

λ(1)n λ

(2)n . . . λ

(n)n

⎞⎟⎟⎟⎠⎛⎜⎜⎜⎝

λ(1) 0 . . . 00 λ(2) . . . 0...

.... . .

...0 0 . . . λ(n)

⎞⎟⎟⎟⎠ . (1.94)

22

Simple inspection shows that T can be constructed as a matrix, where thecolumns are formed by the eigenvectors of M , and D is a diagonal matrixwith the eigenvalues as diagonal elements,

T =(�λ(1) �λ(2) . . . �λ(n)

)⎛⎜⎜⎜⎝

λ(1)1 λ

(2)1 . . . λ

(n)1

λ(1)2 λ

(2)2 . . . λ

(n)2

......

. . ....

λ(1)n λ

(2)n . . . λ

(n)n

⎞⎟⎟⎟⎠ and D =

⎛⎜⎜⎜⎝

λ(1) 0 . . . 00 λ(2) . . . 0...

.... . .

...0 0 . . . λ(n)

⎞⎟⎟⎟⎠ .(1.95)

Example:

• In order to see how this works, take one of the previous examples, cf.Eq. (1.83):

5x(1) − 2y(1) = 6x(1) 5x(2) − 2y(2) = x(2)

−2x(1) + 2y(1) = 6y(1) −2x(2) + 2y(2) = y(2) ,(1.96)

which can be written as one matrix equation, namely(5 −2−2 2

)(x1 x2

y1 y2

)=

(x1 x2

y1 y2

)(1 00 6

). (1.97)

Clearly, multiplying from the left with a suitable inverse,(x1 x2

y1 y2

)−1 (5 −2−2 2

)(x1 x2

y1 y2

)

=

(x1 x2

y1 y2

)−1 (x1 x2

y1 y2

)(1 00 6

)=

(1 00 6

)(1.98)

yields the similarity transformation that transforms the original matrixinto a diagonal from with the eigenvalues on the diagonal.

In general, arbitrary n × n matrices A, may be brought into diagonal formby a similarity transformation

S−1AS = Adiag. , (1.99)

23

where the matrix S corresponding to the similarity transformation is justgiven by the matrix whose rows are given by the eigenvectors of the originalmatrix A and the diagonal matrix is determined by the eigenvalues of A. Itmust be stressed, however, that it is not guaranteed that any matrix A maybe diagonalized.In any case, once a matrix can be diagonalized, it is straightforward to seethat its determinant equals the product of its eigenvalues. In addition, dueto the properties of the similarity transformation diagonalizing the matrix,the trace of a matrix equals the sum of its eigenvalues.

Symmetric matrices and diagonalization

In this section, it will be proved that symmetric matrices can alwaysbe diagonalized and that eigenvectors of different eigenvalues arealways orthogonal.Proof: Consider the eigenvalue equation A�λ = λ�λ in component notation,

Aijλj = λλi . (1.100)

Due to the symmetry property of A, the eigenvalue equation can also beapplied in the form

�κT A = κ�κT or κiAij = κκj (1.101)

Multiplying the first equation from the left with �κT and the second with �λform the right then yields, in components

κiAijλj = λκiλi and κiAijλj = κκjλj . (1.102)

Replacing the summation over j with a summation over i on the r.h.s. of thesecond equation and subtracting both equations results in

0 = κiλi(λ− κ) = �κ · �λ(λ− κ) . (1.103)

This can be zero only, if either he two eigenvalues are identical or if the twovectors are orthogonal, proving the theorem.It is worth noting that a similar theorem also holds true for Hermitian ma-trices: The eigenvalues of Hermitian matrices are real, and eigenvec-tors corresponding to different eigenvalues are orthogonal.

24

Physical applications of diagonalization

• Consider a body M which is made of a material with a constant massdensity ρ. Its inertial tensor is a matrix given by

mij =

∫M

d3rρ(�r)(r2δij − xixj) . (1.104)

This is a diagonal matrix and hence can be diagonalized. It plays animportant role in rotations.

Since the angular momentum and velocity are related by

�L = M�ω (1.105)

this shows that they are parallel only if they point along one of theeigenvectors of the inertial tensor. If, in contrast, they are not parallelto one of these vectors, in general, ω will not be constant but rathermove in some precession or similar.

• Consider next an one-dimensional system where two masses are con-nected by one spring each to a fixed wall and by one spring with eachother. Assume the masses and the spring constants to be identical.Then, calling the excursion from rest of the two masses by x1 and x2

the systems Lagrangian is given by

L = T − V

=m

2(x2

1 + x22) −

k

2[x2 + y2 + (x− y)2] . (1.106)

This can be written ion matrix form as

L = + (x1, x2)

(m/2 0

0 m/2

)(x1

x2

)

− (x1, x2)

(2k/2 −k/2−k/2 2k/2

)(x1

x2

). (1.107)

The potential can be diagonalized with the eigenvalues

λ(1) = 1 and λ(2) = 3 . (1.108)

Now, whatever the similarity transformation is that diagonalizes thematrix of the potential, it is clear that it also leaves the matrix of the

25

kinetic energy in diagonal form, since it is proportional to the unit ma-trix. Denoting the corresponding eigenvectors by �x and �y, the equationsof motion therefore read

�x+ ω2�x = 0 and �y + 3ω2�y = 0 , (1.109)

where ω2 = k/m. This immediately leads to two oscillation modes withdifferent frequency. The eigenvectors are given by

�x =1√2(x1 − x2) and �y =

1√2(x1 + x2) , (1.110)

i.e. a slow oscillations, where both masses are oscillating back and forthin parallel, and a fast oscillation, where the two masses oscillate againsteach other.

26

Chapter 2

Fourier Series

2.1 Motivation

2.1.1 Simple harmonic motion

Many problems in physics are related to oscillations/vibrations. Examplesinclude the pendulum or harmonic oscillators, waves of sound or light, ACelectric currents, etc.. Other phenomena, such as heat expansion, the physicsof electric and magnetic fields, are, not so obvious, described through similarequations. In general, these equations contain a second derivative, somethinglike d2/dt2 for the harmonic oscillator or, in three dimensions, the Laplaceoperator, ∇2. Of course, in order to refer to physical situations, such equa-tions are supplemented with boundary conditions. More than often, however,the best way to solve such equations is through an expansion in harmonicfunctions - a Fourier expansion.Many ideas and a good part of the terminology used in this part of the lecturewill be borrowed from the discussion of simple harmonic motion and wavemotion. Hence, to begin with, these two topics will briefly be reviewed.Suppose a particle P moves with constant speed around a circle with radiusR in the xy-plane. At the same time, let another particle, Q, move on they-axis between −R and R, such that the y-coordinates of P and R coincide atall times. Such a motion is called simple harmonic motion, as exhibited,for instance, by a harmonic oscillator without friction.If the angular velocity of P is ω, and if P started at ϑ(t = 0) = 0, then the

27

y-coordinate of P and Q read

y(t) = R sin(ωt) , (2.1)

and the x-coordinate of P is given by

x(t) = R cos(ωt) . (2.2)

In complex coordinates, one could also write the position of P as

z(t) = x(t) + iy(t)

R exp(iωt) = R cos(ωt) + iR sin(ωt) . (2.3)

It is often worthwhile to use the complex notation also for Q, understandingthat only the real or imaginary component are relevant for the discussion ofits motion.In any case, the velocity of Q is given by the derivative w.r.t. the time t:

v(t) = y(t) = Rω cosωt (2.4)

or the imaginary component of

z(t) = Aiωeiωt . (2.5)

What do these quantities translate to? To see this, it is worthwhile to drawthe function y(t) in dependence of t. Quite obviously this function repeatsitself, it is periodic (see below) with a period of 2π/ω. Physically, it showsthe displacement of Q from y = 0, its equilibrium position. R, the maximaldisplacement, is called the amplitude of the vibration or the displacement.It is also called the amplitude of the function. Similarly, Rω is the maximalvelocity, or the velocity amplitude, of the motion. Note that the velocity hasthe same period as the displacement, it is just shifted by some angle - thedifference of sin and cos. Now, if the mass of Q is m, the kinetic energy isgiven by

Ekin =m

2y2 =

mR2ω2

2cos2(ωt) , (2.6)

and the maximal kinetic energy is given by

Emaxkin =

mR2ω2

2. (2.7)

For an idealized harmonic oscillator which does not use energy due to frictionthe total energy is always a constant, equal to the largest value of the kineticenergy. It is interesting to note that the total energy is proportional to thesquare of the velocity amplitude, waves behave in a similar manner.

28

2.1.2 Waves

Waves are yet another example of an oscillation. To illustrate this, considerhighly idealized (“unrealistic” would put it better) water waves, following asine curve. Taking a photo of the water surface at some time t = 0 thenwould reveal a function

z(x, t = 0) = A sin2πx

λ, (2.8)

where z represents the height of the water surface w.r.t. the equilibrium, xis the horizontal distance in one dimension, and λ is the distance betweenwave crests. Usually, λ is called the wave length of the waves.Suppose now that another photo is taken at some time t �= 0. Assuming thewaves to travel with velocity v in positive x direction, the equation repre-senting the water surface now reads

z(x, t) = A sin2π(x− vt)

λ. (2.9)

There’s another interpretation to this: Rather than waiting for a time t andobserve the change of the water surface at some fixed point x, one may alsofix the time t and consider changes under displacement along x. In otherwords, the equation for z(t) above yields the function of the water surface independence of both x and t and it is a mere question of circumstance whichaspect (t fixed, x variable or vice versa) is considered.

2.1.3 More frequencies

To carry this discussion further, consider a string instrument, such as a violin.Again in extreme idealization, sound is produced in such instruments bydisplacing a string and letting it go again. If there was no energy loss, again,the string would continue to vibrate forever. But what are the allowed modesof this vibration? The answer is quite simple. Assume that the string has alength L. In equilibrium, it stretches along the x-coordinate, and it is fixedat the positions x = 0 and x = L. Displacements in the, say, y direction,can then be described in an oversimplified way as standing waves y(x) of thestring; but the boundary conditions are that y(0) = y(L) = 0. Functionswhich satisfy this are sinnπx/L with n = 1, 2, 3, dots. It is not hard toguess that in this oversimplified world all relevant oscillations of the string

29

can be described as a superposition of these elementary modes. It is exactlythis transformation of arbitrary oscillations into sums of eigenmodes whichmakes Fourier analysis such an important tool.

2.2 Trigonometric and Exponential functions

2.2.1 General properties of functions

Periodic functions

A function f(x) is called periodic with a period P > 0, if

f(x+ P ) = f(x) ∀ x , (2.10)

where P is a constant. The smallest value of P is called the least period orjust the period of f .

Examples:

• the function sin x has periods 2π, 4π, . . . , since, sin(x+ 2π) = sin(x+4π) = · · · = sin x. Obviously, 2π is the least period of sin x;

• the functions sin(nx) and cos(nx) with a constant n have periods 2π/n;

• the function exp(ikx) with a real constant k and with real values of xhas period 2π/k;

• the period of tanx is π;

• constants have any positive number as period.

Piecewise continuous functions

A function f(x) is called piecewise continuous in an interval x ∈ I, if

1. the interval can be divided into a finite number of subintervals, in whichf(x) is continuous; and

30

2. the limits of f(x) are finite as x approaches the subintervals’ endpoints.

Another, more pictorial way, of stating these conditions is to demand thatf(x) can be patched together with a finite number of patches with only finitejumps in between them.The limit of f(x) from the right is often denoted as

lim f(x+ 0) = limε→0+

f(x+ ε) , (2.11)

where ε > 0. Similarly, the left limit is denoted as

lim f(x− 0) = limε→0+

f(x− ε) = limε→0−

f(x+ ε) . (2.12)

Odd and even functions

A function is called odd, if f(−x) = −f(x). If, on the other hand f(−x) =f(x) then the function is called even.

Examples include:

• polynomials with even exponents only, such as a0 + a2x2 + a4x

4 + . . . ,where the ai are constants, are even. In contrast, polynomials with oddexponents only, such as a1x+ a3x

3 + a5x5 + . . . are odd functions;

• sin x is an odd function, whereas cosx is even.

It is quite simple to show that the product of two odd or two even functionsis even, whereas the product of an odd and an even function is odd. Also,

A∫−A

dx f(x) =

⎧⎨⎩

0 if f(x) is odd ,

2A∫0

dx f(x) if f(x) is even .(2.13)

2.2.2 Some trigonometry

Euler’s identity

Euler’s identity states that

eiϑ = cosϑ+ i sinϑ . (2.14)

31

This can be seen geometrically or using the Taylor expansion

f(x) = f(x0) + (x− x0)d

dxf(x)

∣∣∣∣x=x0

+(x− x0)

2

2

d2

dx 2f(x)

∣∣∣∣x=x0

+ . . .

=

∞∑n=0

(x− x0)n

n!

dn

dx nf(x)

∣∣∣∣x=x0

, (2.15)

where the obvious definition d0/dx 0f(x) = f(x) and 0! = 1 have been used.Applying this on the functions above with ϑ0 = 0 yields

eiϑ = 1 + iϑ− 1

2ϑ2 − 1

6iϑ3 +

1

24ϑ4 +

i

120ϑ5 + . . .

cos(ϑ) = 1 − 1

2ϑ2 +

1

24ϑ4 + . . .

i sin(ϑ) = iϑ − i

6ϑ3 +

i

120ϑ5 + . . . . (2.16)

Euler’s identity above together with the properties sin(−ϑ) = − sin ϑ andcos(−ϑ) = cos ϑ of the trigonometric functions implies that

e−iϑ = cos ϑ− i sinϑ (2.17)

and thus

cosϑ =eiϑ + e−iϑ

2

sinϑ =eiϑ − e−iϑ

2i. (2.18)

Trigonometric functions: addition & subtraction

From the identities above, Eq. (2.18), it is easy to see that

sin(2ϑ) =e2iϑ − e−2iϑ

2i

=1

2i

[(eiϑ

)2 − (e−iϑ

)2]

=1

2i

[(cos ϑ+ i sin ϑ)2 − (cosϑ− i sinϑ)2]

=1

2i[4i(cosϑ sin ϑ] = 2 cosϑ sin ϑ . (2.19)

32

In a similar way,

sinϑ+ sinφ =eiϑ − e−iϑ + eiφ − e−iφ

2i

=1

2i

[exp

(iϑ+ φ

2+ i

ϑ− φ

2

)− exp

(−iϑ+ φ

2− i

ϑ− φ

2

)

+ exp

(iφ+ ϑ

2+ i

φ− ϑ

2

)− exp

(−iφ + ϑ

2− i

φ− ϑ

2

)]

=1

2i

[exp

(iϑ+ φ

2

)− exp

(−iϑ + φ

2

)]

×[exp

(iϑ− φ

2

)+ exp

(−iϑ− φ

2

)]

= 2 sin(ϑ+ φ) cos(ϑ− φ) . (2.20)

Also,

sin ϑ− sinφ = 2 cos(ϑ+ φ) sin(ϑ− φ)

cosϑ+ cosφ = 2 cos(ϑ+ φ) cos(ϑ− φ)

cosϑ− cosφ = −2 sin(ϑ+ φ) sin(ϑ− φ) . (2.21)

Trigonometric functions: products

Euler’s identity can also be employed to express products of trigonometricfunctions by sums and differences. For instance,

sinϑ sinφ =eiϑ − e−iϑ

2i

eiφ − e−iφ

2i

= −1

4

[ei(ϑ+φ) − ei(ϑ−φ) − e−i(ϑ−φ) + e−i(ϑ+φ)

]=

1

2[cos(ϑ− φ) − cos(ϑ+ φ)] . (2.22)

Also,

sin ϑ cosφ =1

2[sin(ϑ− φ) + sin(ϑ+ φ)] =

1

2[sin(φ+ ϑ) − sin(φ− ϑ)]

cosϑ cosφ =1

2[cos(ϑ− φ) + cos(ϑ+ φ)] . (2.23)

33

In particular, this implies for the squares of sine or cosine functions that

sin2 ϑ =1

2(1 − cos 2ϑ)

cos2 ϑ =1

2(1 + cos 2ϑ) . (2.24)

Trigonometric functions: derivatives and integrals

Remember that

d sin(kx)

dx= +k cos(x) and

d cos(kx)

dx= −k sin(x) , (2.25)

where k is a constant.Therefore, their definite integrals read

b∫a

dx sin(kx) = −cos kb− cos ka

k

b∫a

dx cos(kx) = +sin kb− sin ka

k. (2.26)

Choosing k = 1 and b− a = 2π immediately shows that the integrals of thetrigonometric functions over 2π vanish1.For the integrals of the squared trigonometric functions, this however doesnot hold true. There are different ways of calculating, say, the integral ofsin2(x). In particular, one of them is well worth knowing. It starts byrealizing that

sin2 x+ cos2 x = 1 . (2.27)

Also, when integrated over the full period, both sin2 x and cos2 x yield the

1It is important to distinguish between definite and indefinite integrals: The formeryields numbers as result, whereas the latter yield functions.

34

same result,

2π∫0

dx sin2 x =

2π∫0

dx cos2 x =

2π∫0

dx (1 − sin2 x)

=⇒2π∫0

dx = 2π = 2

2π∫0

dx sin2 x

=⇒ π =

2π∫0

dx sin2 x =

2π∫0

dx cos2 x . (2.28)

This immediately implies that the average value of sin2 x, denoted by 〈sin2 x〉,is 1/2:

〈sin2 x〉 =1

2π

2π∫0

dx sin2 x =π

2π. (2.29)

Trigonometric functions: integrals of products

Finally, let us calculate the integral (average value) of products of trigono-metric functions over a period. To start with, consider

2π∫0

dx sin(nx) cos(mx) =1

2

2π∫0

dx {sin[(n+m)x] + sin[(n−m)x]} = 0 ,

(2.30)

where the identities of Eq. (2.23) have been used. With the same procedure,

2π∫0

dx sin(nx) sin(mx) =

⎧⎨⎩

0 for n �= mπ for n = m �= 00 for n = m = 0

2π∫0

dx cos(nx) cos(mx) =

⎧⎨⎩

0 for n �= mπ for n = m �= 02π for n = m = 0

(2.31)

can be obtained.

35

2.3 Definition & convergence

2.3.1 Definition

Assume a function f(x) with the following properties:

• f(x) is defined in the interval [L, −L];

• f(x) is periodic with period 2L, i.e., f(x+ 2L) = f(x).

Then, the Fourier series (or Fourier series expansion) of f(x), here denotedas f(x), is defined through

f(x) =a0

2+

∞∑n=1

(an cos

nπx

L+ bn sin

nπx

L

)(2.32)

where the coefficients (also known as Fourier coefficients) are given by

an =1

L

L∫−L

dx f(x) cosnπx

L

bn =1

L

L∫−L

dx f(x) sinnπx

L. (2.33)

A motivation of the special form of these coefficients will be given in the nextsection, cf. Sec. .This immediately gives an interpretation for the first term in the series ex-pansion, a0/2. Using the form of the Fourier coefficients above, it is clearthat this term is nothing but the mean of f(x) in the interval [−L, L],

a0

2=

1

2L

L∫−L

dx f(x) cos0πx

L=

1

2L

L∫−L

dx f(x) .

(2.34)

In writing down these expressions for the Fourier expressions, it has implicitlybeen assumed that the function in question is continuous. If this is not the

36

case, the coefficients are obtained through integration over the continuoussubintervals and adding the results.It must be emphasized, however, that the Fourier series is only a series thatcorresponds to f(x). It is not clear a priori (and in fact, counter examplescan be constructed), whether it converges or, even if it does so, whether itconverges to f(x).

Examples:

• Consider first a constant, like f(x) = c in the interval [−π, π]. Re-peating itself with period 2π of course yields a constant function forall values of x. The Fourier coefficients are then given just by therespective integrals,

a0

2=

c

2π

π∫−π

dx = c ;

an =c

π

π∫−π

dx cos(nx) =c

π

sin(nx)

n

∣∣∣∣π

−π= 0 ;

bn =c

π

π∫−π

dx sin(nx) = − c

π

cos(nx)

n

∣∣∣∣π

−π= 0 , (2.35)

and the Fourier series related to this function is the function itself.

• Consider now the function f(x) = xΘ(x) in the interval [−π, π], withperiod 2π. Then

a0

2=

1

π

π∫0

dxx =π

4;

an =1

π

π∫0

dxx cos(nx) =1

π

x sin(nx)

n

∣∣∣∣π

0

− 1

π

π∫0

dxsin(nx)

n

=1

π

cos(nx)

n2

∣∣∣∣π

0

=1

πn2[(−1)n − 1] =

⎧⎨⎩

−2

n2πfor n odd

0 for n even ;

37

bn =1

π

π∫0

dxx sin(nx) = −1

π

x cos(nx)

n

∣∣∣∣π

0

+1

π

π∫0

dxcos(nx)

n

= −1

π

(x cos(nx)

n+

sin(nx)

n2

)∣∣∣∣π

0

= −(−1)n

n. (2.36)

In all cases, integration by parts has been used. Taken together, theFourier series then reads

f(x) =π

4+

∞∑n=1

[−1 − (−1)n

πn2cos(nx) − (−1)n

nsin(nx)

].(2.37)

• As a next example, take the function f(x) = x in the interval [0, π],repeating itself with period π. Obviously, the interval is not symmetricaround 0, but this merely changes the region of integration and the co-efficients of the sine and cosine functions. Then, the Fourier coefficientsare given by

a0

2=

1

π

π∫0

dxx =π

2;

an =2

π

π∫0

dxx cos(2nx) =2

π

x sin(2nx)

2n

∣∣∣∣π

0

− 2

π

π∫0

dxsin(2nx)

2n

=2

π

cos(nx)

n2

∣∣∣∣π

0

= 0

bn =2

π

π∫0

dxx sin(nx) = −2

π

x cos(2nx)

2n

∣∣∣∣π

0

+2

π

π∫0

dxcos(2nx)

2n

= −2

π

(x cos(nx)

2n+

sin(2nx)

(2n)2

)∣∣∣∣π

0

= −1

n. (2.38)

The related Fourier series thus reads

f(x) =π

2−

∞∑n=1

sin(2nx)

n. (2.39)

38

• Now, take the function f(x) = x2 in the interval [−π, π], extendedperiodically with period 2π. The Fourier coefficients read

a0 =2π2

3

an =4(−1)n

n2

bn = 0 . (2.40)

Thus, the Fourier series is given by

f(x) =π2

3+ 4

∞∑n=1

(−1)n

n2cos(nx) . (2.41)

It will be left as a problem to check this.

• Consider the Heavyside (or step) function

Θ(x) :=

{0 for x ≤ 01 for x > 0

(2.42)

in the interval [−π, π] and repeat it periodically with period 2π. ItsFourier coefficients read

a0

2=

1

2π

π∫−π

dxΘ(x) =1

2;

an =1

π

π∫−π

dxΘ(x) cos(nx) =1

π

π∫0

dx cos(nx) =sin(nx)

πn

∣∣∣∣π

0

= 0 ;

bn =1

π

π∫−π

dxΘ(x) sin(nx) =1

π

π∫0

dx sin(nx) = − cos(nx)

nπ

∣∣∣∣π

0

=

⎧⎨⎩

2

nπfor n odd

0 for n even .(2.43)

Hence, the Fourier series related to the step function reads

f(x) =1

2+

∞∑n=1

1

(2n− 1)πsin((2n− 1)x)

=1

2+

2

π

[sin(x) +

sin(3x)

3+

sin(5x)

5+ . . .

]. (2.44)

39

Note that for x = 0 this yields f(0) = 1/2, quite in contrast to thevalue of f(x = 0) = Θ(0) = 0.

2.3.2 The form of the Fourier coefficients

To convince oneself that the form of the Fourier coefficients, cf. Eq. (2.33, isindeed correct, consider the Fourier series of a function f(x) (Eq. (2.32)),

f(x) =a0

2+

∞∑n=1

(an cos

nπx

L+ bn sin

nπx

L

), (2.45)

multiply it with cos(mπx/L) and integrate over x in the interval [−L, L].This yields

L∫−L

dx f(x) cosmπx

L

=

L∫−L

dx cosmπx

L

[a0

2+

∞∑n=1

(an cos

nπx

L+ bn sin

nπx

L

)]

=La0

2mπsin

mπx

L

∣∣∣∣L

−L+

∞∑n=1

⎡⎣an

⎛⎝ L∫−L

dx cosnπx

Lcos

mπx

L

⎞⎠

+ bn

⎛⎝ L∫−L

dx sinnπx

Lcos

mπx

L

⎞⎠⎤⎦

=

∞∑n=1

anLδnm = amL , (2.46)

where in the last step the properties of Sec. 2.2.2 have been employed. Thisimmediately yields

am =1

L

L∫−L

dx f(x) cosmπx

L. (2.47)

40

Similar statements of course hold true also for the coefficients bn. In otherwords, assuming the Fourier series to coincide with the original function,i.e. f(x) = f(x), fixes the expansion coefficients. So, again, the questionremains, when the Fourier expansion coincides with the original function.This is answered in the next section.

2.3.3 Convergence: Dirichlet conditions

The Dirichlet conditions below answer the question, whether the Fourierseries converges at all.Suppose f(x) fulfils the following conditions:

1. f(x) is defined and single valued in the interval [−L, L] except at afinite number of values;

2. f(x) has a period of 2L;

3. both f(x) and its first derivative f ′(x) are piecewise continuous in[−L, L].

Then the Fourier series converges such that

f(x) −→{f(x) if x is a point of continuity;f(x+ 0) + f(x− 0)

2if x is a point of discontinuity.

(2.48)

An example for such a point of discontinuity is x = 0 for the Heavysidefunction, Θ(x). As could be seen from Eq. (2.44), its Fourier series thereyields 1/2, in accordance with the above statements.

The three conditions above are sufficient but not necessary: it is possibleto have convergence without some of them. However, if they are met bythe function f(x), one may substitute f(x) with the series expansion at thefunction’s points of continuity. In any case, it is interesting to note thatcontinuity of f alone does not guarantee the convergence of the series andthat rather also its first derivative must be piecewise continuous.

41

2.3.4 Uniform convergence of series

Suppose an infinite series

u(x) =∞∑n=1

un(x) . (2.49)

The Rth partial sum SR(x) of the series is then defined as the sum of the Rfirst terms,

SR(x) =

R∑n=1

un(x) . (2.50)

Now, by definition, the series u(x) converges to a function u(x) in someinterval, if for any value of x in the interval and for any positive number εthere exists a positive number N , such that

|SR(x) − u(x)| < ε ∀ R > N . (2.51)

In general, this number N depends on both ε and the possible values of x.Anyways, u(x) will be called the sum of the series in the following.An important case, namely uniform convergence, occurs, when N does notdepend on the range of possible x but on ε only. Such uniform convergentseries have two important properties, namely

1. If each term of an infinite series (i.e. each of the un(x)) is continuous inan interval ]a, b[ and if the series u(x) is uniformly convergent to thesum u(x), then

• u(x) is also continuous in the interval;

• the series can be integrated term by term in this interval,

b∫a

dx

( ∞∑n=1

un(x)

)=

∞∑n=1

⎛⎝ b∫

a

dxun(x)

⎞⎠ . (2.52)

In other words, in this case, summation and integration commute- they are interchangeable operations.

42

2. If, in addition, each term of this series has a derivative and if theseries of derivatives is uniformly convergent, then the series can bedifferentiated term by term,

d

dx

( ∞∑n=1

un(x)

)=

∞∑n=1

(dun(x)

dx

). (2.53)

2.3.5 The Weierstrass M-test

There are various ways to prove uniform convergence, the most obvious beingto actually find SR(x) in closed form and then apply the definition directly.Another, more powerful way is to use the Weierstrass M-test:If there exists a set of constants Mn with n = 1, 2, 3, dots such that

|un(x)| ≤ Mn ∀x ∈ [a, b] (2.54)

and if furthermore the sum∞∑n=1

Mn converges, then

u(x) =∞∑n=1

un(x) (2.55)

converges uniformly in the interval [a, b]. In fact, in this case, u(x) is abso-lutely convergent, i.e. also the sum

∞∑n=1

|un(x)| (2.56)

of the absolute values of the terms is convergent.

Example: Consider the series∞∑n=1

sin(nx)n2 . It converges uniformly in [−π, π]

(in fact, everywhere), because a set of constants Mn = 1/n2 can be found,such that ∣∣∣∣sin(nx)

n2

∣∣∣∣ ≤ 1

n2and

∞∑n=1

1

n2=π2

6. (2.57)

43

2.4 Simple properties

2.4.1 Parseval’s identity

Parseval’s identity states that, if a function f(x) satisfies the Dirichletconditions,

1

L

L∫−L

dx [f(x)]2 =a2

0

2+

∞∑n=1

(a2n + b2n) , (2.58)

where the an and bn are the Fourier coefficients defined in Eq. (2.33), whichcorrespond to f(x).The proof of this identity will be left as a problem.

2.4.2 Sine and Cosine series

Expanding even and off functions

Remember the properties of even and odd functions, when multiplied witheach other, cf. Sec. 2.2.1. If the function f(x) is an even function in theinterval [L, −L] (i.e. f(−x) = f(x)) and periodic with period 2L, its Fouriertransform will contain only the cos-terms. This is quite simple to see: Aneven function (like f(x)) when multiplied with an odd function (like sin(kx))yields an odd function. And odd function vanish, when integrated over asymmetric interval like [−L, L]. Hence, in this case, all coefficients bn equal0, and thus all sin terms vanish. In this case, the Fourier series reduces to aCosine series.On the other hand, if f(x) is an odd function in the interval, all its Fouriercoefficients an equal 0, and thus the Fourier series reduces to a Sine series.

Examples:

• Consider first the function

f(x) = x (2.59)

in the interval [−π, π] and re-express is as either a Sine or Cosine series.Clearly, this is an odd function, therefore it can be expected that allcoefficients an vanish.

44

From the definition of the Fourier coefficients, Eq. (2.33), one finds

a0

2=

1

2π

π∫−π

dx f(x) =1

2π

π∫−π

dxx =x2

4π

∣∣∣∣π

−π= 0 ;

an =1

π

π∫−π

dx f(x) cos(nx) =1

π

π∫−π

dxx cos(nx)

=1

π

x sin(nx)

n

∣∣∣∣π

−π− 1

π

π∫−π

dxsin(nx)

n=

1

π

cos(nx)

n2

∣∣∣∣π

−π= 0 ;

bn =1

π

π∫−π

dx f(x) sin(nx) =1

π

π∫−π

dxx sin(nx)

= −1

π

x cos(nx)

n

∣∣∣∣π

−π+

1

π

π∫−π

dxcos(nx)

n= −2(−1)n

n. (2.60)

Hence the Fourier expansion of f(x) = x looks like

f(x) = −2∞∑n=1

(−1)n

nsin(nx)

= 2

[sin(x) − sin(2x)

2+

sin(3x)

3+ . . .

]. (2.61)

• On the other hand, consider the function

f(x) = |x| (2.62)

in the interval [−π, π] and re-express is as either a Sine or Cosine series.Since this function is even, it can be expected that now all coefficientsbn vanish.

From the definition of the Fourier coefficients, Eq. (2.33), one finds

a0

2=

1

2π

π∫−π

dx f(x) =1

2π

π∫−π

dx |x| = − 1

2π

0∫−π

dxx+1

2π

π∫0

dxx

= − x2

4π

∣∣∣∣0

−π− x2

4π

∣∣∣∣π

0

=π

2;

45

an =1

π

π∫−π

dx f(x) cos(nx) = −1

π

0∫−π

dxx cos(nx) +1

π

π∫0

dxx cos(nx)

= − x sin(nx)

nπ

∣∣∣∣0

−π−

0∫−π

dxsin(nx)

nπ− x sin(nx)

nπ

∣∣∣∣π

0

+

π∫0

dxsin(nx)

nπ

= − cos(nx)

n2π

∣∣∣∣0

−π+

cos(nx)

n2π

∣∣∣∣π

0

=2(cosnπ − 1)

n2π

=

{− 4

n2πif n is odd

0 if n is even ;

bn =1

π

π∫−π

dx f(x) sin(nx) = −1

π

0∫−π

dxx sin(nx) +1

π

π∫0

dxx sin(nx)

=x cos(nx)

nπ

∣∣∣∣0

−π−

0∫−π

dxcos(nx)

nπ− x cos(nx)

nπ

∣∣∣∣π

0

+

π∫0

dxcos(nx)

nπ

=(−1)nπ

nπ− (−1)nπ

nπ− sin(nx)

n2π

∣∣∣∣0

−π+

sin(nx)

n2π

∣∣∣∣π

0

= 0 . (2.63)

Hence, in this case, the Fourier transform of f(x) looks like

f(x) =π

2+

4

π

∞∑n=0

cos[(2n+ 1)x]

(2n+ 1)2

=π

2+

4

π

[cos(x) +

cos(3x)

9+

cos(5x)

25+ . . .

]. (2.64)

Half-range expansions

Assume now that a function is given as periodic over a range [−L, L], butspecified only in the interval [0, L]. In this case, one may use the freedomof defining the function in the other half of the period in a way such that

46

its Fourier expansion can be written entirely as a sine- or cosine series. Thetrick here is to define the function in [−L, 0] such that it, taken over the fullperiod, f(x) is either even or odd.Due to the symmetry properties of even and odd functions, f±

ext(x), beingmultiplied with cosine or sine functions one finds

{anbn

}=

1

L

L∫−L

dxf±ext(x)

{cosnxsinnx

}=

2

L

L∫0

dxf(x)

{cosnxsinnx

}, (2.65)

potentially alleviating the task of calculating the Fourier coefficients.

Examples:

• Consider first the function f(x) = x in [0, 2] with period 4. Anodd expansion is to define f−

ext(x) = x, whereas an even expansionis f+

ext(x) = |x|. In both cases, the Fourier expansion reads

f−ext = −4

π

∞∑n=1

(−1)n

nsin

nπx

2

=4

π

(sin πx2 − 1

2sin 2πx2 +

1

3sin 3πx2 − 1

4sin 4πx2 + . . .

)(2.66)

and

f+ext = 1 +

4

π2

∞∑n=1

(−1)n − 1

n2cos

nπx

2

= 1 − 8

π2

(cosπx2 +

1

32cos 3πx2 +

1

52cos 5πx2 + . . .

),

(2.67)

respectively.

• Maybe even more instructive is the expansion of the function f(x) =sin(x), x ∈ [0, π] in terms of a cosine series. To this end the full 2πperiod is covered by defining f(x)+

ext = | sin x|, i.e. by extending the

47

half-range sine to an even full-range function. Then

a0

2=

1

π

π∫0

dx sin x =2

π,

an =2

π

π∫0

dx sin x cos(nx) =1

π

π∫0

dx [sin(x+ nx) + sin(x− nx)]

=1

π

[−cos[(n + 1)x])

n+ 1+

cos[(n− 1)x]

n− 1

]π0

= −2[1 + (−1)n]

π(n2 − 1)if n �= 1 ,

a1 =2

π

π∫0

dx sin x cos x =2

π

sin2 x

2

∣∣∣∣π

0

= 0 . (2.68)

Therefore,

f+ext(x) =

2

π

[1 −

∞∑n=2

1 + (−1)n

n2 − 1cosnx

]

=2

π− 4

π

[cos 2x

22 − 1+

cos 4x

42 − 1+

cos 6x

62 − 1+ . . .

]. (2.69)

2.4.3 Integration and differentiation

Integration and differentiation of Fourier series can be justified through thetheorems of Sec. 2.3.4. In other words: If the Fourier series is uniformlyconvergent in its interval [−L, L] it can be integrated term by term. Hence,

48

in this case

L∫−L

dx f(x) =

L∫−L

dx

[a0

2+

∞∑n=1

(an cos

nπx

L+ bn sin

nπx

L

)]

=a0

2

L∫−L

dx +∞∑n=1

⎡⎣ L∫−L

dx(an cos

nπx

L+ bn sin

nπx

L

)⎤⎦

=a0x

2

∣∣∣L−L

+∞∑n=1

⎡⎣an

L∫−L

dx cosnπx

L+ bn

L∫−L

dx sinnπx

L

⎤⎦

= a0L+

∞∑n=1

[anL

nπsin

nπx

L− bnL

nπcos

nπx

L

]L−L

= a0L . (2.70)

This shows that, if the Fourier series is converging to the original function, theintegral of both, function and Fourier transform, over the full period coincideand equal the first term in the Fourier series (the mean of the function) timesthe size of the interval. This should not come as a surprise at this stage.

Example: Consider the Fourier expansion of the function f(x) = x, cf. Eq.(2.61),

f(x) = 2

[sin(x) − sin(2x)

2+

sin(3x)

3+ . . .

]. (2.71)

Integrating this yields∫dx f(x) = 2

∫dx

[sin(x) − sin(2x)

2+

sin(3x)

3+ . . .

]

= 2

[− cos(x) +

cos(2x)

4− cos(3x)

9+ . . .

]= 2

n∑n=1

(−1)n cos(nx)

n2.

(2.72)

Since f(x) satisfies the Dirichlet conditions, f(x) can be replaced by f(x)

49

and thus

x2

2= 2

n∑n=1

(−1)n cos(nx)

n2

∣∣∣∣x

0

= 2

n∑n=1

(−1)n cos(nx)

n2− 2

n∑n=1

(−1)n

n2. (2.73)

Adding in that

n∑n=1

(−1)n

n2= 1 − 1

4+

1

9− 1

16+ · · · =

π2

12(2.74)

yields

x2 =π2

3+ 4

n∑n=1

(−1)n cos(nx)

n2, (2.75)

where the r.h.s. is exactly the Fourier expansion for x2, cf. Eq. (2.41.It must be stressed here however that in general the term-by-term integra-tion of a Fourier series does not reproduce the Fourier expansion of the corre-sponding integrated function; this is due to the presence of the a0-term whichresults in something like a0L, where L is the length of the integration interval.

If in addition, also the derivatives are uniformly convergent, the series canalso be differentiated term by term. But it is by no means guaranteed thatthe differentiated series has anything to do with the derivative of the originalfunction.

Example: As an example for this consider again

x = 2

[sin(x) − sin(2x)

2+

sin(3x)

3+ . . .

], (2.76)

cf. Eq. (2.61). Differentiating on both sides w.r.t. x clearly yields 1 for thel.h.s.. The right hand side, however, reads

2 [cos(x) − cos(2x) + cos(3x) + . . . ] , (2.77)

50

clearly different from 1. Even worse, this series does not converge at all.

Nevertheless, if both the function itself, f(x) and its derivative, f ′(x), satisfythe Dirichlet conditions in the interval in question then term-by-term differ-entiation of the Fourier series of f(x) is permissible in the sense that it leadsto the Fourier series of f ′(x).The catch here is that the Dirichlet conditions demand continuity of thefunction, even when expanded with its given periodicity. Simple inspectionshows that for f(x) = x2 in the interval [−π, π] this holds true, basicallybecause f(π) = f(−π). In contrast for the function f(x) = x in the sameinterval, there are discontinuities at the boundaries of the period.

2.4.4 Complex notation

Another form of the Fourier series can be obtained by using Euler’s relation

eiϑ = cosϑ+ i sinϑ (2.78)

and its inverse

cosϑ =eiϑ + e−iϑ

2and sin ϑ =

eiϑ − e−iϑ

2i. (2.79)

These can easily be inserted into the Fourier expansion and, also, in the ex-pressions determining the Fourier coefficients. But, of course, the coefficientscan also be found directly and, using Euler’s equation, the original sine andcosine terms can be recovered.It is instructive to see how the coefficients of the complex Fourier seriesexpansion emerge. To that end, consider a series of the form

f(x) = c0 + c1eix + c−1e

−ix + c2e2ix + c−2e

−2ix + · · · =∞∑

n=−∞cne

inx (2.80)

of a function f(x) satisfying Dirichlet’s conditions in the interval [−π, π].Direct calculation (and previous experience with the trigonometric functions)shows that indeed

π∫−π

dx eikx = 0 (2.81)

51

for all integer values of k different from 0. Hence, integrating over the periodyields

π∫−π

dx f(x) = 2πc0 (2.82)

and thus

c0 =1

2π

π∫−π

dx f(x) , (2.83)

the average value of the function in the interval.To find any of the cn the recipe clearly is to compensate for the exp(inx) termon the r.h.s. of the equation above. This is most easily done by multiplyingwith exp(−inx) before integrating. Thus

π∫−π

dx f(x)e−inx =

π∫−π

dx cneinxe−inx = 2πcn (2.84)

and therefore

cn =1

2π

π∫−π

dx f(x)e−inx (2.85)

for all values of n.Note that this equation automatically contains the c0 term as well, there areno factors of 1/2 to be concerned about. Also,

c−n = c∗n if f(x) is real . (2.86)

Example: Consider, once more, the Heavyside function in the interval [−π, π].

52

Its complex Fourier expansion reads

c0 =1

2π

π∫−π

dxΘ(x) =1

2

cn =1

2π

π∫−π

dxΘ(x)e−inx

=1

2π

π∫0

dx e−inx = −e−inx

2inπ

∣∣∣∣π

0

=1 − einı

2inπ=

⎧⎨⎩

1

inπfor n odd

0 for n even and n �= 0 .(2.87)

Hence,

f(x) =1

2+

2

π

[eix − e−ix

2i+

1

3

e3ix − e−3ix

2i+

1

5

e5ix − e−5ix

2i+ . . .

]

=1

2+

2

π

[sin x+

sin 3x

3+

sin 5x

5+ . . .

], (2.88)

as before, cf. Eq. (2.44).

2.4.5 Double and multiple Fourier series

It is straightforward to extent the idea of Fourier expansion of a functionof one variable to functions with more variables. As an example consider afunction f(x, y) defined in {x, y} ∈ [−π, π] and periodic with period 2π ineach dimension. Such a function can be expanded in a double Fourier series,like e.g.

f(x, y) =

∞∑n=−∞

∞∑m=−∞

cnmeinxeimy =

∞∑n,m=−∞

cnmeinx+imx , (2.89)

53

where the coefficients are given by

cnm =1

(2π)2

π∫−π

dx

π∫−π

dyf(x, y)e−inx−imy . (2.90)

2.5 Some applications

2.5.1 Full wave rectifier

Wave rectifiers translate alternating into direct current (AC→DC). An inter-esting question related to this is how well the resulting current approachesdirect current. As a toy example consider a wave rectifier, which lets positiveparts of a sine wave pass and which inverts the negative parts,

sin(ωt)Rect.−→ f(t) =

{+ sin(ωt) for 0 ≤ ωt < π− sin(ωt) for − π ≤ ωt < 0 .

(2.91)

Since the result is an even function, there won’t be any sin-parts in theFourier series of the result. Consequently, only the an need to be calculated,

a0

2= − 1

2π

0∫−π

d(ωt) sin(ωt) +1

2π

π∫0

d(ωt) sin(ωt) =2

π;

an = −1

π

0∫−π

d(ωt) sin(ωt) cos(nωt) +1

π

π∫0

d(ωt) sin(ωt) cos(nωt)

=

⎧⎨⎩ − 4

(n2 − 1)πfor n even

0 for n odd .(2.92)

Hence, the resulting series is

f(t) =2

π− 4

π

∞∑n=2,4,6...

cos(nωt)

n2 − 1=

2

π

[1 − 2 cos(2ωt)

3− 2 cos(3ωt)

8− . . .

],

(2.93)

which is clearly dominated by the first few terms. Also, the original frequencyω has been eliminated and replaced by its higher harmonics 2ω, 3ω etc..

54

2.5.2 Quantum mechanical particle in a box

Sketching the physical problem

A good example for the merit of applying Fourier expansions, consider thequantum mechanical problem of a particle in a box. Such problems are oftendescribed by the time-independent Schrodinger equation. In one dimension,it reads

�2

2m

d2

dx2uE(x) = [V (x) − E]uE(x) , (2.94)

where the uE(x) are the energy eigenfunctions. They are related to the fullwave function by

ψ(x, t) =∑E

e−iEtuE(x) (2.95)

with the sums stretching over all allowed energy eigenvalues of the problem.The box can now be represented by a potential of the form

V (x) =

{0 for x ∈ [−a, a]V else ,

(2.96)

with V = ∞. This infinitely high potential confines the particle to the regionbetween −a and a, it cannot penetrate the potential walls. According to theprobabilistic interpretation of the wave function, its value

ψ(x, t) = e−iE/�tuE(x) = 0 for x /∈ [−a, a] (2.97)

translates into the particle being in the region. Therefore, suitable boundaryconditions for the wave function inside the region [−a, a] are

uE(−a) = uE(a) = 0 . (2.98)

General solution of the differential equation

A suitable ansatz for its solution is

uE(x) = B+ exp(ikx) +B− exp(−ikx) , (2.99)

where

k =

√2mE

�2. (2.100)

55

Plugging this into the Schrodinger equation, translates into

− �2

2m(ik)2 [B+ exp(ikx) +B− exp(−ikx)]

= E[B+ exp(ikx) +B− exp(−ikx)] (2.101)

in the region x ∈ [−a, a], proving the validity of the ansatz.

Adding in boundary conditions

Adding in boundary conditions

uE(±a) = B+ exp(±ika) +B−(∓ika) = 0 (2.102)

implies there are two sets of solutions, namely

• Sine-type solutions:

B+ = −B− ≡ A− and k =2nπ

2a

=⇒ uE(x) = A−[exp

(2niπx

2a

)− exp

(−2niπx

2a

)](2.103)

• Cosine-type solutions:

B+ = B− ≡ A+ and k =(2n− 1)π

2a

=⇒ uE(x) = A+

[exp

((2n− 1)iπx

2a

)+ exp

(−(2n− 1)iπx

2a

)](2.104)

In both cases, n = 1, 2, 3, . . . .

Brief discussion of the physics

These two sets of solutions, +- and −-types, correspond to sinoidal andcosinoidal solutions and, ultimately, to different parities. Here, parity is theproperty of an object under spatial reflection, i.e. under the transformation�x → −�x, or, in one dimension, x → −x. More precisely, parity is definedthrough

O �x→−�x−→ PO . (2.105)

56

Then clearly, the ±-parts have parity P = ±. Thus the ±-label also labelsthe parity of the respective eigenmode.The corresponding energy eigenmodes have kinetic energies only (the poten-tial vanishes), namely

E−n =

�2k2

2m=

�2

2m

(2n)2π2

(2a)2

E+n =

�2k2

2m=

�2

2m

(2n− 1)2π2

(2a)2. (2.106)

Hence, each individual mode can also be written as

u−En(x) = 2A−

n sin2nπx

2a

u+En

(x) = 2A+n cos

(2n− 1)πx

2a. (2.107)

To fix the amplitudes, it must be taken into account that the energy eigen-modes are also normalized, implying

1 ≡a∫

−a

dx|u±En(x)|2 = 4a|A±n |2 (2.108)

leading to

An =1

2√a

(2.109)

Taken everything together yields

u±n (x) =1√a

⎧⎪⎨⎪⎩

cos(2n− 1)πx

2a

sin2nπx

2a.

(2.110)

for each allowed value n ∈ N. This can easily be generalized to the case ofmore than one dimension, i.e. a three-dimensional box with dimensions a, band c.

57

2.5.3 Heat expansion

Sketching the physical problem

Heat conduction can be described by

∂u(�x, t)

∂t= κ∇2u(�x, t) , (2.111)

where the function u(�x, t) denotes the temperature at position �x at a giventime t, and the constant κ is called diffusity and is related to

κ =K

σμ, (2.112)

where K is the thermal conductivity, σ is the specific heat, and μ denotesthe density of the material in question. The expression ∇2 is called theLaplacian, given by

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2. (2.113)

To see how this works, consider the one-dimensional case of a slab with lengthL. Both its ends and the entire surface are insulated, translating into no heattransferred beyond the edges of the slab. In mathematical form this can beexpressed as

u′(0, t) = u′(L, t) = 0 . (2.114)

Assume also that the slab has some initial temperature, say at time t = 0,then

u(x, 0) = f(x) ., (2.115)

General solution of the differential equation

Let us see now, how Fourier expansion helps us to determine the subsequenttemperature, i.e. u(x, t) with t > 0. The starting point will be an inves-tigation of the general anatomy of the solutions of the differential equationdescribing heat conduction. For this, the ansatz

u(x, t) = X(x)T (t) (2.116)

58

will be employed, leading to

XT ′ = κX ′′T (2.117)

when put into the equation of heat conduction, Eq. (2.111) . This can bere-expressed as

T ′

κT=X ′′

X. (2.118)

Since T is a function of t only and X merely depends on x, it should be clearthat the equality can be realized only, if both sides equal a constant for allvalues of t and x, respectively. Setting this constant to be −λ2, therefore

T ′ + κλ2T = 0 and X ′′ + λ2X = 0 . (2.119)

Solutions for these two equations are given by

T = C exp(−κλ2t) and X = A cos(λx) +B sin(λx) , (2.120)

yielding

u(x , t) = exp(−κλ2t) [a cos(λx) + b sin(λx)] , (2.121)

with the constants a = AC and b = BC.

Adding in boundary conditions

These constants as well as the constant −λ2 need to be determined from theboundary conditions:

• The condition u′(0, t) = 0 enforces all sine terms to vanish, becausetheir derivative would yield a cosine, which does not vanish for itsargument being 0. Hence

b = 0 . (2.122)

• The condition u′(L, t) = 0 thus translates into

cos′(λL) ∝ sin(λL)!= 0 =⇒ λL = mπ rm with m = 1, 2, 3, . . .(2.123)

and therefore

λ =mπ

L. (2.124)

59

• Finally, the boundary condition u(x, 0) = f(x) must be met. Hence

u(x, 0) = f(x) =a0

2+

∞∑m=1

am cos(mπx

L

), (2.125)

the result for t = 0. It looks like a Fourier cosine series, with coefficientsthus given by

a0

2=

1

L

L∫0

dx f(x)

am =1

L

L∫0

dx f(x) sin(mπx

L

), (2.126)

cf. Eq. (2.33).

Hence,

u(x, t) =1

L

L∫0

dx f(x)

+∞∑m=1

⎧⎨⎩⎡⎣ 1

L

L∫0

dx f(x) sin(mπx

L

)⎤⎦ · cos(mπx

L

)exp

(−κm

2π2t

L2

)⎫⎬⎭ .

(2.127)

2.5.4 Electrostatic potential

Sketching the physical problem

In the previous example, it has been seen that partial differential equationscan sometimes be solved by separating the variables. By using this method,one often encounters sets of orthogonal functions, which are of great impor-tance by themselves. In the following, this method will be applied once more,this time to the case of electrostatics. In the static case, the electric field �Eis determined through Gauss’ law. In differential form it is given by

ε0�∇ �E = ρ , (2.128)

60

where ρ(x) is the fixed charge density giving rise to the field. A nice wayof solving this problem is to realize that the rotor of the electric field in thestatic case vanishes, allowing to introduce the electrostatic potential φ. Itrelates to the electric field through

�E = −�∇φ . (2.129)

The potential is thus determined through the inhomogeneous Laplace equa-tion

∇2φ = −ε0ρ , (2.130)

encountered before. This equation is of great importance in physics.

Homogeneous Laplace equation

Let us now consider a case, where the potential φ(�x) can be written in Carte-sian coordinates as a product of three functions, one for each coordinate,

φ(�x) = X(x)Y (y)Z(z) . (2.131)

Setting the charge density to zero for a moment, the (homogeneous) Laplaceequation reads

1

X

d2X

d2x+

1

Y

d2Y

d2y+

1

Z

d2Z

d2z= 0 . (2.132)

Here the partial differentials became total ones, since the functions dependon one variable only. In order to guarantee a solution for all �x, the followingsystem has to be solved

1

X

d2X

d2x= α2

1

Y

d2Y

d2y= β2

1

Z

d2Z

d2z= γ2 , (2.133)

with the constraint that

α2 + β2 + γ2 = 0 . (2.134)

A general solution for the homogeneous Laplace equation therefore reads

φ(�x) = φ0e±αxe±βye±i

√α2+β2z , (2.135)

with φ0 the potential at �x = 0.

61

A physical problem and its solution

In order to determine α and β, boundary conditions need to be specified.As an example consider a cuboid with dimensions a × b × c in x-, y, andz-direction, respectively. Furthermore, assume that all sides are on potentialzero, apart from the one at z = c, which has a fixed potential Vz=c(x, y).What is the potential inside the cuboid? The constraint that φ = 0 forx = 0, y = 0, z = 0 implies

X = sin(αx) , Y = sin(βy) , Z = sinh(√α2 + β2z) . (2.136)

The additional constraint that φ = 0 for x = a, y = b yields

αa = nπ and βb = mπ , (2.137)

where n and m are integers. From these identification of the frequencies, theexpansion coefficients of the Z function can be inferred. They read

γnm =√α2 + β2 = π

√n2

a2+m2

b2. (2.138)

Thus, the potential can be written as a sum of components, where each ofwhich is given by

φnm = Anm sin(nπx

a

)sin

(mπyb

)sinh (γnmz) . (2.139)

By construction this potential vanishes for x = 0, a, y = 0, b, and for z = 0.The coefficients Anm can now be fixed such that∑

nm

φnm(x, y, z = c) = Vz=c(x, y) . (2.140)

This seems horrible at first. By taking a closer look, however, it becomesclear that it is just a double Fourier series for the function Vz=c. Thus, thecoefficients Anm are given by

Anm =1

sinh (γnmc)· 4

ab

a∫0

dx

b∫0

dyVz=c(x, y) sin(nπx

a

)sin

(mπyb

).

(2.141)

If the potential is different from zero on all sides of the cuboid, then theoverall potential can be constructed from the superposition principle: Justrepeat this procedure for the other five sides, and add the partial results.

62

Chapter 3

Fourier transforms

3.1 Definitions and properties

3.1.1 The Fourier integral

The need for Fourier integrals

So far, Fourier series have been defined for periodic functions with period L.A natural question to ask is what happens, if L → ∞. It will turn out thatin such cases, the Fourier series becomes a Fourier integral, wherethe sum over discrete frequencies is replaced by an integral over acontinuum of frequencies. This Fourier integral may be used to representalso non-periodic functions, such as a single voltage pulse or a flash of light,and it also represents contiuous spectra of frequencies, like whole ranges oftones or colours. In that respect it is often used for the description of a widerange of physical phenomena.In addition, especially in Quantum Mechanics, it is used to connect thedescription of physical problems in position space with the correspondingdescription in momentum space.

Definition

To keep things short at this stage, let us begin by defining the Fourier integral.Fourier’s integral theorem states that functions f(x) enjoing the followingproperties

• f(x) and f ′(x) are piecewise continuous in any finite interval, and

63

• the integral of f(x) is absolutely convergent, i.e.

∞∫−∞

dx|f(x)| = M , where |M | <∞ (3.1)

can be represented as

f(x) =

∞∫0

dk [A(k) cos(kx) +B(k) sin(kx)] . (3.2)

Here, the coefficent functions read

A(k) =1

π

∞∫−∞

dx f(x) cos(kx)

B(k) =1

π

∞∫−∞

dx f(x) sin(kx) . (3.3)

It must be stressed, however, that the representation above holds true onlyfor points of continuity of f(x). If, in contrast, f(x) is discontinuous at apoint x, then the representation above will yield

limy→x+

f(y) + limy→x−

f(y)

2=

∞∫0

dk [A(k) cos(kx) +B(k) sin(kx)] . (3.4)

The similarity of the Fourier integral defined here with the Fourier seriesconsiderred in the previous chapter is apparent - this is why more than oftenthe term Fourier expansion is used for both the discrete series and for theintegral presented here.

Equivalent forms & a simple motivation

There is a number of equivalent forms for the above Fourier integral, one ofwhich will be discussed below. As a side-product, some further motivationfor the Fourier integral and its specific form will emerge. It should be kept inmind, however, that the following is just a sketch of a more formal derivation.

64

The starting point of the reasoning is the Fourier series of a function f(x),

f(x) =1

2L

L∫−L

dz f(z) +1

L

∞∑m=1

cosmπx

L

⎡⎣ L∫−L

dzf(z) cosmπz

L

⎤⎦

+1

L

∞∑m=1

sinmπx

L

⎡⎣ L∫−L

dzf(z) sinmπz

L

⎤⎦ .

(3.5)

Combining two of the equations in Eq. (2.21), namely

cosϑ+ cosφ = 2 cos(ϑ+ φ) cos(ϑ− φ)

cos ϑ− cosφ = −2 sin(ϑ+ φ) sin(ϑ− φ) (3.6)

yields

cosα cosβ + sinα sin β = cos(α− β) (3.7)

and thus

f(x) =1

2L

L∫−L

dz f(z) +1

L

∞∑m=1

⎡⎣ L∫−L

dzf(z) cosmπ(x− z)

L

⎤⎦ . (3.8)

Consider now the case, where L→ ∞. Setting

k =mπ

Land Δk =

π

L(3.9)

and using

cosα =eiα + e−iα

2(3.10)

yields

f(x) =1

2L

L∫−L

dz f(z) cosmπ(x− z)

L

∣∣∣∣m=0

+1

2L

∞∑m=1

⎡⎣ L∫−L

dzf(z)

(cos

imπ(x− z)

L+ cos

−imπ(x− z)

L

)⎤⎦ ,

65

allowing to write

f(x) =1

2L

∞∑m=0

⎡⎣ L∫−L

dzf(z)

(exp

imπ(x− z)

L+ exp

−imπ(x− z)

L

)⎤⎦ .

In other words, the sum stretches now from 0 rather than from 1 to infinity.Then, transforming the exponentials back to cosines,

f(x) −→∞∑m=0

Δk

π

∞∫−∞

dzf(z) cos[k(x− z)]

Δk→0−→ 1

π

∞∫0

dk

∞∫−∞

dzf(z) cos[k(x− z)] . (3.11)

Employing symmetry arguments, i.e. cos(−x) = cosx, this can be cast into

f(x) =1

2π

∞∫−∞

dk

∞∫−∞

dzf(z) cos[k(x− z)] . (3.12)

Using again the identity

cos(α− β) = cosα cosβ + sinα sin β (3.13)

yields the desired result.It is straightforward to convince oneself that application of the same relationsof sine and cosine function will yield the original form of the Fourier integralabove.

Complex form

At this point it is worthwhile to notice that the integrand above, f(z) cos[k(x−z)] is an even function of k and, hence, yields a non-vanishing contributionwhen integrated over k. In contrast, f(z) sin[k(x− z)] is an odd function ink, yielding a zero result after integration over k. Therefore, there’s no harm

66

in adding, say, f(z)i sin[k(x− z)] to the integrand above. Thus,

f(x) =1

2π

∞∫−∞

dk

∞∫−∞

dzf(z) cos[k(x− z)]

=1

2π

∞∫−∞

dk

∞∫−∞

dzf(z) {cos[k(x− z)] + i sin[k(x− z)]}

=1

2π

∞∫−∞

dk

∞∫−∞

dzf(z) exp[ik(x− z)] =1

2π

∞∫−∞

dkeikx∞∫

−∞

dzf(z)e−ikz .

(3.14)

Thus, in its complex form, the Fourier transform can be written as

f(x) =1

2π

∞∫−∞

dkeikx∞∫

−∞

dzf(z)e−ikz ≡ 1

2π

∞∫−∞

dkeikxF (k) .

(3.15)

Due to its definition, the Fourier transform

F (k) =

∞∫−∞

dzf(z)e−ikz (3.16)

enjoys the following property

F (−k) = F ∗(k) , (3.17)

where the ∗ denotes complex conjugation, i.e. the operation i→ −i.

3.1.2 Detour: The δ-function

Let’s go back to the derivation above, specifically, consider

f(x) =1

2π

∞∫−∞

dk

∞∫−∞

dzf(z) exp[ik(x− z)]

=

∞∫−∞

dz f(z)

⎧⎨⎩ 1

2π

∞∫−∞

dk eik(x−z)

⎫⎬⎭ (3.18)

67

Apparently, the expression in the curly brackets depends on the difference z−x; morevore it obviously projects out the value of f(z = x) in the remaining zintegral. This is exactly the behaviour of Dirac’s δ-function (which in fact formathematicians is a distribution rather than a function). It has the propertythat

∞∫−∞

dz f(z)δ(x− z) = f(x) , (3.19)

if f is continuous at the point x. Anyway, written in this fashion, the waythe δ-function behaves becomes more obvious: it acts under integration inthe same way a Kronecker-δ does under summation.To properly define the δ function:

b∫a

dx f(x)δ(x− y) =

{f(y) if y ∈ [a, b]0 else .

(3.20)

Representations

The form given above in the curly brackets serves as one way (among others)of representing this δ-function:

δ(x) =1

2π

∞∫−∞

dk eikx . (3.21)

It must be stressed, however, that anything said here about this function isvalid only under an integration over x.There’s yet another way to look at it. To this end, consider a sequence ofδn(z − x) defined through

δn(z − x) =sin[n(z − x)]

π(z − x)=

1

2π

n∫−n

dk eik(z−x) . (3.22)

With some effort, it can be shown that

f(x) = limn→∞

∞∫−∞

dz f(z)δn(z − x) . (3.23)

68

Taking this as a starting point, inserting the definition of the δn from above,and reshuffling the order of integration yields

f(x) =1

2πlimn→∞

∞∫−∞

dz f(z)

n∫−n

dk eik(z−x)

=1

2πlimn→∞

n∫−n

dk

∞∫−∞

dz f(z)eik(z−x) =1

2π

∞∫−∞

dke−ikx∞∫

−∞

dz f(z)eikz ,

(3.24)

exactly the form of the Fourier integral encountered in Eq. (3.14).

Important properties

The first property, concerns the integral over a product of a function and thederivative of the δ-function. Understanding the δ function with the help ofthe representation above, it is clear that partial integration can be applied,leading to ∫

dxf(x)δ′(x− a) = −f ′(a) . (3.25)

If the argument of the δ-function itself is again a function, say g(x), then

∞∫−∞

dx f(x)δ(g(x)) =∑xi

∣∣∣∣∣[dg(x)

dx

]x=xi

∣∣∣∣∣−1

f(xi) , (3.26)

where the xi are the (simple) zeroes of g(x): g(xi) = 0. In particular,

∞∫−∞

dx f(x)δ(x2 − a2) =f(x− a) + f(x+ a)

2|a| . (3.27)

In more dimensions, the δ-function is just the product of δ-functions in thecorresponding dimensions:

δ(�x) = δ(x)δ(y)δ(z) . (3.28)

69

3.1.3 Fourier transforms

Fourier transforms and their inverse

In the previous section the Fourier integral has been discussed; in its complexform it has been given by

f(x) =

∞∫−∞

dk F (k)eikx , (3.29)

where the coefficient function F (k) has been defined through

F (k) =1

2π

∞∫−∞

dx f(x)e−ikx . (3.30)

In order to pronounce the symmetry of the two forms, the normalisationfactor 1/2π neccessary to go from arguments x to the arguments k and back,can be re-distributed. The weay how to that is purely conventional, anychoice will do, as long as one sticks to it. A natural choice is to distribute itevenly between the two:

f(x) =1√2π

∞∫−∞

dk F (k)eikx

F (k) =1√2π

∞∫−∞

dx f(x)e−ikx . (3.31)

This is actually the choice that will be used in the following.In any case, the function F (k) is known as the Fourier transform of f(x),and, vice versa, f(x) is the Fourier transform of F (k).

A cool example

Consider the function

f(x) =

{1 for |x| < a0 for |x| > a ,

(3.32)

70

where a > 0. Its Fourier transform is

F (k) =1√2π

∞∫−∞

dxf(x)e−ikx

=1√2π

a∫−a

dxe−ikx =1√2π

e−ikx

−ik∣∣∣∣a

−a

=1√2π

2 sin(ka)

kif k �= 0 . (3.33)

For k = 0, F (k) = 2a/√

2π can be obtained by taking the limit and applyingthe rule of l’Hopital.So far, so good. But how can this result be used for something more interest-ing? After all, the title of this paragraph is “A cool example”. To see this,remember the inverse Fourier transform. It connects the integral

f(x) =1√2π

∞∫−∞

dkF (k)eikx

=1

2π

∞∫−∞

dk2 sin(ka)

keikx =

⎧⎨⎩

1 for |x| < a1/2 for |x| = a0 for |x| > a ,

(3.34)

with a finite value. The catch here is that there seems to be a pole at k = 0.To proceed, let us go back to the inverse Fourier transform and replace theexponential first with sines and cosines:

f(x) =1

2π

∞∫−∞

dk2 sin(ka)

keikx =

1

2π

∞∫−∞

dk2 sin(ka)

k[cos(kx) + i sin(kx)]

=1

2π

∞∫−∞

dk2 sin(ka) cos(kx)

k(3.35)

because the integrand in the second term is an odd function of k. This impliesthat

∞∫−∞

dksin(ka) cos(kx)

k=

⎧⎨⎩

π for |x| < aπ/2 for |x| = a0 for |x| > a .

(3.36)

71

Choosing now x = a, using

sin(ka) cos(ka) =1

2[sin(ka− ka) + sin(ka+ ka)] =

sin(2ka)

2(3.37)

and employing the symmetry of the integral implies that

π =

∞∫0

dksin(ka) cos(ka)

k=

∞∫0

dksin(2ka)

2k(3.38)

and hence

π

2=

∞∫0

dksin(ka)

k. (3.39)

So far, a > 0 has been assumed. Similar reasoning shows that for a < 0, thevalue of the integral merely changes sign and therefore

∞∫0

dksin(ka)

k=

{+π

2for a > 0

−π2

for a < 0

}=π

2[Θ(a) − Θ(−a)] . (3.40)

Fourier Sine and Cosine transforms

Similar to the case of Fourier series, it is tempting to use the symmetry prop-erties of the function f(x) in order to alleviate the Fourier transformation;the symmetry in question is, of course, the parity of the function.For even or odd functions f±(x) (i.e. functions with positive or negativeparity), it is sufficient to consider Fourier Cosine or Fourier Sine Trans-forms, respectively. They are given by

f+(x) =1√2π

∞∫−∞

dk F+(k) cos(kx) =2√2π

∞∫0

dk F+(k) cos(kx)

F+(k) =1√2π

∞∫−∞

dk f+(x) cos(kx) =2√2π

∞∫0

dk f+(x) cos(kx)

(3.41)

72

and

f−(x) =1√2π

∞∫−∞

dk F−(k) sin(kx) =2√2π

∞∫0

dk F−(k) sin(kx)

F−(k) =1√2π

∞∫−∞

dk f−(x) sin(kx) =2√2π

∞∫0

dk f−(x) sin(kx)

(3.42)

However, there are cases, discussed below, where Fourier Sine and CosineTransforms may be sensible independent of the parity of the function.

Example:

• The Fourier Cosine Transform of

f(x) = e−mx , with m > 0 (3.43)

is given by

Fc(k) = F+(k) =

√2

π

∞∫0

dxe−mx cos(kx)

=1√2π

∞∫0

dxe−mx[eikx + e−ikx]

= − 1√2π

[e−(m−ik)x

m− ik+e−(m+ik)x

m+ ik

]∞0

=

√2

π

m

m2 + k2. (3.44)

This can be used to show that∞∫

0

dkcos(kx)

m2 + k2=

√π

2

1

m

√2

π

∞∫0

dkm cos(kx)

m2 + k2=

π

2me−mx ,

(3.45)

where in the last transformation the inverse Fourier Transformationhas been used.

With similar tricks a large number of integrals involving trigonometricfunctions can be solved.

73

Parseval’s identity, once again

In Sec. 2.4.1 Parseval’s identity for Fourier series has been discussed. In thissection, an analogous expression for Fourier transforms will be presented:If F (k) and G(k) are Fourier transforms of f(x) and g(x), respectively, then

∞∫−∞

dxf(x)g(x)

=

∞∫−∞

dx

⎧⎨⎩⎡⎣ 1√

2π

∞∫−∞

dkF (k)eikx

⎤⎦⎡⎣ 1√

2π

∞∫−∞

dk′G(k′)eik′x

⎤⎦⎫⎬⎭

=1

2π

∞∫−∞

dkdk′F (k)G(k′)

∞∫−∞

dxei(k+k′)x .

At this point, it is worthwhile to remember the representations of the δ-function to yield

∞∫−∞

dxf(x)g(x) =1

2π

∞∫−∞

dkdk′F (k)G(k′)δ(k + k′)

=1

2π

∞∫−∞

dkF (k)G(−k) =1

2π

∞∫−∞

dkF (k)G∗(k) , (3.46)

where the ∗ denotes the complex conjugate, obtained by replacing i with −i.In particular, if f(x) = g(x) is a real function, then

∞∫−∞

dx|f(x)|2 =1

2π

∞∫−∞

dk|F (k)|2 . (3.47)

74

Convolution theorem

In a similar fashion, the convolution theorem can be proved. A convolutionof two functions, (f ∗ g)(x) is defined through

(f ∗ g)(x) ≡∞∫

−∞

dyf(y)g(x− y) . (3.48)

The Fourier Transform of such a convolution reads

(F ∗G)(k) = (2π)F (k)G(k) . (3.49)

To see this consider

(2π)F (k)G(k) =2π

2π

∞∫−∞

dyeikyf(y)

∞∫−∞

dzeikzg(z)

=

∞∫−∞

dydzeik(y+z)f(y)g(z) (3.50)

and transform from y, z to x = y + z, y. Then

(2π)F (k)G(k) =

∞∫−∞

dxdyeikxf(y)g(x− y)

=

∞∫−∞

dxeikx

⎡⎣ ∞∫−∞

dyf(y)g(x− y)

⎤⎦

=

∞∫−∞

dxeikx(f ∗ g)(x) = (F ∗G)(k) . (3.51)

In other words, up to a factor of 2π the Fourier Transform of a convolu-tion of two functions equals the product of their Fourier Transforms takenindividually,

(F ∗G)(k) = 2πF (k)G(k) . (3.52)

75

3.2 Applications

3.2.1 Finite wave train

An important application of Fourier transforms is the resolution of a finitepulse into a continous spectrum of sinusoidal waves. To highlight this, imag-ine an infinite wave train with frequency ω0 is cut such that

f(t) =

{sin(ω0t) for |t| < nπ

ω0

0 else .(3.53)

This corresponds to cutting out n cyxcles of the original wave train.Since, obviously, f(t) is odd, it is sufficient to analyse its Fourier sine trans-form, given by

F−(ω) =

√2

π

∞∫0

dt sin(ωt)f(t)

=

√2

π

nπ/ω0∫0

dt sin(ωt) sin(ω0t)

=1√2π

nπ/ω0∫0

dt [cos(ω − ω0)t− cos(ω + ω0)t]

=1√2π

⎡⎢⎢⎣

sinnπ(ω − ω0)

ω0

ω − ω0−

sinnπ(ω + ω0)

ω0

ω + ω0

⎤⎥⎥⎦ . (3.54)

It is quite interesting, how this depends on frequency: For ω = ω0 + δ and δsmall compared to ω and ω0, the first term dominates. Then

F−(ω) ≈ 1√2πδ

sinnπδ

ω, (3.55)

yielding zeroes for

δ = ω − ω0 = ±ωn, ±2ω

n, . . . . (3.56)

76

In fact, this corresponds very well to the amplitude curve for the single slitdiffraction pattern, exhibiting the same zeroes. It is worth noting that themaximum is given for δ = 0, i.e. for a resonance. In this case,

F−max(ω) =

nπ√2πω

. (3.57)

In any case, it is tempting to take the dfistance to the first zero as a measurefor the spread of the wave pulse. It is given by

Δω =ω

n(3.58)

and is small for large trains and large for small trains.

3.2.2 Infinite string

Sketching the physical problem

Consider an infinitely long string, stretched along the x-axis. Displacements,say in y-direction at one point of this string will then propagate along it,such that they satisfy a wave equation with

1

α2

∂2y

∂t2=∂2y

∂x2. (3.59)

Here, α plays the role of an inverse wave velocity.A simple way to initiate a wave on this string is to just displace it at sometime t = 0 according to a function y(x, 0) = f(x) with ∂y(x, 0)/∂t = 0. Itmakes perfect sense to assume that this initial displacement is of only limitedamplitude, i.e. |y(x, t) < M .

General solution

To find a general solution, let’s again use a separation ansatz,

y(x, t) = X(x)T (t) =[A+(k)eikx + A−(k)e−ikx

] [B+(ω)eiωt +B−(ω)e−iωt

],(3.60)

where, in principle the coefficients A± and B± are functions of k and ω, re-spectively. However, plugging this into the differential equation above yields

ω2

α2= k2 . (3.61)

77

In other words,

1

α=k

ω. (3.62)

Since k and ω are connected by a mere constant, the dependence on theformer can easily be replaced by a dependence on the latter (or vice versa).

Including boundary conditions

The boundary conditions read

v(x, 0) = f(x) =[A+(k)eikx + A−(k)e−ikx

] [B+(k) +B−(k)

]vt(x, 0) = 0 =

[A+(k)eikx + A−(k)e−ikx

]iω

[B+(k) − B−(k)

],(3.63)

yielding

B+(k) = B−(k) = B(k) (3.64)

and, renaming A±(k) → A±(k) = B(k)A±(k) yields

f(x) = A+(k)eikx + A−(k)e−ikx = X(x) (3.65)

and something like

y(x, t) = f(x) cos(kαt) . (3.66)

This already looks very nice, but there’s even more that can be gained. Inorder to do so, let us Fourier expand f(x):

f(x) =1√2π

∞∫−∞

dk [C(k) cos(kx) + S(k) sin(kx)] (3.67)

with

C(k) =1√2π

∞∫−∞

dxf(x) cos(kx)

S(k) =1√2π

∞∫−∞

dxf(x) sin(kx) . (3.68)

78

Putting everything together yields

y(x, t) =1

2π

∞∫−∞

dk

∞∫−∞

dx′f(x′) [cos(kx) cos(kx′) + sin(kx) sin(kx′)] cos(kαt) .

(3.69)

Employing, once more,

cosα cos β + sinα sin β = cos(α− β) , (3.70)

this yields

y(x, t) =1

2π

∞∫−∞

dk

∞∫−∞

dx′f(x′) cos[k(x− x′)] cos(kαt)

=1

4π

∞∫−∞

dk

∞∫−∞

dx′f(x′) {cos[k(x− x′ + αt)] + cos[k(x− x′ − αt)]} ,

(3.71)

where in the last step

cosα cosβ =1

2[cos(α− β) + cos(α + β)] (3.72)

has been used. Changing the order of integration results in

y(x, t) =1

4π

∞∫−∞

dx′∞∫

−∞

dkf(x′) cos[k(x− x′ + αt)]

+1

4π

∞∫−∞

dx′∞∫

−∞

dkf(x′) cos[k(x− x′ − αt)]

=1

2f(x+ αt) +

1

2f(x− αt) . (3.73)

This last line shows that, if the string is distorted for instance by a δ-like“kick”, this kick will propagate with half the amplitude to the left and theright.

79

3.3 Green functions

Green functions are extremely useful in the solution of partial differentialequations, but of course they can also be used in order to solve ordinarydifferential equations.

3.3.1 Green function for the driven harmonic oscillator

What is the Green function?

This will be discussed employing the example of a driven harmonic oscillator.Ignoring damping, the driven harmonic oscialltor is described by

y′′ + ω2y = f(t) . (3.74)

Let us discuss the case where the oscillator initially is at rest, i.e.

y(0) = y′(0) = 0 . (3.75)

Since the driving force f(t) can be rewritten as a sequence of “kicks” through

f(t) =

∞∫0

dt′f(t′)δ(t− t′) , (3.76)

it is sufficient to construct the most general solution for

y′′ + ω2y = δ(t− t′) (3.77)

instead. Denoting this solution by G(t, t′), i.e.

d2G(t, t′)dt2

+ ω2G(t, t′) =

(d2

dt2+ ω2

)G(t, t′) = δ(t− t′) (3.78)

the true solution y(t) can be obtained by “adding up the kicks”,

y(t) =

∞∫0

dt′G(t, t′)f(t′) . (3.79)

80

In order to see this, insert this form of y(t) into the original differentialequation:

y′′(t) + ω2y(t) =

(d2

dt2+ ω2

) ∞∫0

dt′G(t, t′)f(t′)

=

∞∫0

dt′(

d2

dt2+ ω2

)G(t, t′)f(t′) =

∞∫0

dt′δ(t− t′)f(t′) = f(t) .

(3.80)

The function G(t, t′) is called the Green function of the problem. It is theresponse of the system to a unit impulse at t = t′.It is quite simple to show that

G(t, t′) = G(t− t′) =

{0 for 0 < t < t′

sin[ω(t− t′)]ω

for t′ < t

=sin[ω(t− t′)]

ωΘ(t− t′) (3.81)

is a solution of (d2

dt2+ ω2

)G(t, t′) = δ(t− t′) . (3.82)

For 0 < t < t′ (d2

dt2+ ω2

)G(t, t′) = 0 , (3.83)

because in iths case, G(t− t′) = 0, whereas for t′ < t(d2

dt2+ ω2

)G(t, t′) =

(−ω2 + ω2)G(t, t′) = 0 , (3.84)

as demanded.

Explicit construction of the Green function

But how can such a solution be constructed? In order to do so, let us Fouriertransform the full equation. As has been seen in Sec. 3.1.2, a representation

81

of the δ-function is given by the exponential

δ(t− t′) =1

2π

∞∫−∞

dEeiE(t−t′) =1√2π

∞∫−∞

dEeiE(t−t′)Δ(E) (3.85)

such that its Fourier transform Δ(E) is readily identified with 1/√

2π. Forthe l.h.s. of Eq. (3.78), it is useful to realise first that it must be a functionof t − t′, since also the r.h.s. of this equation is a function of t − t′. Then,Fourier transforming G(t, t′) = G(t− t′) yields

G(t− t′) =1√2π

∞∫−∞

dEeiE(t−t′)G(E) . (3.86)

Plugging this into the equation above results in(−E2 + ω2)G(E) = 1/

√2π (3.87)

and thus

G(E) = − 1√2π

1

E2 − ω2. (3.88)

This now needs to be inserted into the equation for G(t− t′):

G(t− t′) = − 1

(√

2π)2

∞∫−∞

dEeiE(t−t′)

E2 − ω2

= − 1

2π

∞∫−∞

dE1

2ω

[eiE(t−t′)

E − ω− eiE(t−t′)

E + ω

]

= − 1

4πω

∞∫−∞

dy

[eiy(t−t

′)eiω(t−t′)

y− eiy(t−t

′)e−iω(t−t′)

y

]

= −2i sin[ω(t− t′)]4πω

∞∫−∞

dyeiy(t−t

′)

y. (3.89)

82

The remaining integral is, formally speaking, ill-defined. For our purpose,however, it suffices to realise that the integral of the δ-function is the Heavyside-or Θ-function1,

Θ(x) =

∫dxδ(x) . (3.94)

Adding in one of the representations of the δ-function, namely

δ(t− t′) =1

2π

∞∫−∞

dy eiy(t−t′) (3.95)

immediately shows that

i

∞∫−∞

dyeiy(t−t

′)

iy= i

∞∫−∞

dy

∫d(t− t′)eiy(t−t

′)

= 2πi

∫d(t− t′)δ(t− t′) = 2πiΘ(t− t′) , (3.96)

1Alternatively, it can be done by help of residues,

∞∫−∞

dyeiy(t−t′)

y= 2πi

k∑ν=1

Res

[eiy(t−t′)

y; yν

], (3.90)

where the sum stretches over all zeroes of the function (i.e. over the zero at y = 0), andthe residuum is given by

Res [f(x); xν ] =1

(m − 1)!lim

x→xν

dm−1[(x − xν)mf(x)]dxm−1

. (3.91)

In the case considered here,

Res

[eiy(t−t′)

y; 0

]=

10!

limy→0

d0[(y − 0)1 eiy(t−t′)y ]

dx0= lim

y→0eiy(t−t′) = 1 . (3.92)

Then,

∞∫−∞

dyeiy(t−t′)

y= 2πi (3.93)

83

and thus

G(t− t′) = −2i sin[ω(t− t′)]4πω

2πiΘ(t− t′) =sin[ω(t− t′)]

ωΘ(t− t′) , (3.97)

as advertised.

Using the Green function

Let us go back to Eq. (3.79),

y(t) =

∞∫0

dt′G(t, t′)f(t′) =

t∫0

dt′sin[ω(t− t′)]

ωf(t′) (3.98)

after the form of the Green function of the driven harmonic oscillator hasbeen inserted, cf. Eq. (3.81).This enables us to immedaitely write the solution for the driving force

f(t) = sin(ω0t) for t > 0 , (3.99)

where again the boundary conditions y(0) = y′(0) = 0 have been employed.It reads

y(t) =

t∫0

dt′sin[ω(t− t′)]

ωsin(ω0t

′)

=1

2ω

t∫0

dt′ {cos[ω(t− t′) − ω0t′] − cos[ω(t− t′) + ω0t

′]}

=1

2ω

[−sin[ωt− (ω + ω0)t

′]ω + ω0

+sin[ωt− (ω − ω0)t

′]ω − ω0

]t0

=1

2ω

[sin(ω0t) + sin(ωt)

ω + ω0

+sin(ω0t) − sin(ωt)]

ω − ω0

]

=1

ω

⎡⎢⎣sin

(ω + ω0)t

2cos

(ω − ω0)t

2ω + ω0

−sin

(ω − ω0)t

2cos

(ω + ω0)t

2ω − ω0

⎤⎥⎦

(3.100)

It is tedious but simple to prove that this is indeed a solution to the problem.

84